Chapter 1 Some basic concepts of Chemistry Questions and Answers: NCERT Solutions for Class 11th Chemistry (Partnership Chemistry )

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Class 11 Chemistry (Partnership Chemistry ) Chapter 1: Some basic concepts of Chemistry - Questions and Answers of NCERT Book Solutions.

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1. Calculate the molecular mass of the following: i) H2O

Ans: H2O The molecular mass of water, H2O, can be calculated by the following steps given below: = (2 × Atomic mass of hydrogen element) + (1 × Atomic mass of oxygen element) = (2 × 1.0084 u) + (1 × 16.00 u) = 2.016 u + 16.00 u = 18.016 u or 18.02 u
ii) CO2
Ans: CO2 The molecular mass of carbon dioxide,CO2, is calculated below: = (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen) = (1 × 12.011 u) + (2 × 16.00 u) = 12.011 u + 32.00 u = 44.01 u iii) CH4
Ans: CH4 The molecular mass of methane,CH4, is calculated below in step by step manner: = (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen) = (1 × 12.011 u) + (4 × 1.008 u) = 12.011 u +4.032 u = 16.043 u

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2. Calculate the mass percent of different elements present in sodium sulphate

Ans: The given compound in the question is sodium sulphate and its formula is Na2SO4. Its molecular formula is calculated below:
Na2SO4=[(2 × 23.0)+(32.066)+4(16.00)]
= 142.066 g
Now, let us find the mass percentage of each element in the given compound using the formula given below:
Mass percent of an element =Mass of that element in the compound Molar mass of the compound×100
∴Mass percent of sodium:
=46.0 g142.066 g × 100
= 32.379
= 32.4%
Now, let us find the mass percentage of sulphur:
=32.066 g142.066 g × 100
= 22.57
= 22.6%
Now, the mass percentage of oxygen:
=64.0 g142.066 g × 100
= 45.049
= 45.05%

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3. Determine the empirical formula of an oxide of iron which has 69.9%iron and 30.1% oxygen by mass.

Ans: We are given that the percentage of iron by mass is 69.9% and the percentage of oxygen by mass is 30.1%.
Now, we can calculate the relative moles of iron by using the formula given below:
=% of iron by massAtomic mass of iron
=69.955.85
= 1.25
In the same way we can calculate the relative moles of oxygen:
=% of oxygen by massAtomic mass of oxygen<
=30.116.00
= 1.88
Since we have relative moles of both elements, we can calculate the simpler molar ratio.
= 1.25: 1.88
= 1: 1.5
= 2: 3
So, now we can write the empirical formula of iron oxide as Fe2O3.

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4. Calculate the amount of carbon dioxide that could be produced when
i) 1 mole of carbon is burnt in the air.

It is possible to express the balanced response of carbon combustion as follows:
C+O2→CO2
Carbon dioxide is produced when one mole of carbon is burned in one mole of dioxygen (air) using the balancing equation.
ii) 1 mole of carbon is burnt in 16 g of dioxygen.
Ans: It is possible to express the balanced response of carbon combustion as follows:
C+O2→CO2
There is only 16 g of dioxygen accessible, as stated in the question. In this case, it will react with 0.5 moles of carbon to produce 22 g of CO2. In this sense, it is a limiter.
iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Ans: Ans: It is possible to express the balanced response of carbon combustion as follows:
C+O2→CO2
It appears that just 16 g of dioxygen is accessible. In other words, it's a reaction-suppressing agent. This means that with just half as much carbon as dioxygen, 22 g of carbon dioxide may be produced.

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5. Calculate the mass of sodium acetate [Math Processing Error] required to make 500 mLof 0.375 molar aqueous Ans. Molar mass of sodium acetate is

A0.375 M aqueous Ans of sodium acetate means 1000 ml of Ans contains 0.375 moles of sodium acetate.
So, we can calculate the number of moles of sodium acetate in 500 ml. This is given below:
=0.3751000×500
= 0.1875 mole
We are also given the molar mass of sodium acetate is 82.0245 g mole−1.
Required mass of sodium acetate can be calculated as:=(82.0245 g mol−1)(0.1875 mole)
= 15.38 g

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6. Calculate the concentration of nitric acid in moles per liter in a sample which has a density, 1.41 g mL−1 and the mass percent of nitric acid in it being 69%.

Ans: We are given the mass percentage of nitric acid in the sample as 69%.
So, we can say that 100 g of nitric acid contains 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO3) is calculated as below:
= 1 + 14 + (3 × 16)
= 1 + 14 + 48
= 63 g mol−1
As we have the molar mass of nitric acid so, we can find the number of moles in 69 g of nitric acid as given below:
=69 g63 g mol−1
= 1.095 mol
Volume of 100 g of nitric acid And can be calculated as:
=Mass of solutiondensity of solution
=100 g1.41 g mL−1
=70.92 mL
=70.92 × 10-3 L
Now, we can calculate the concentration of nitric acid as:
=1.095 mole70.92 × 10-3 L
= 15.44 mol/ L
Therefore, the concentration of nitric acid is 15.44 mol/ L

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7. How much copper can be obtained from 100 g of copper sulphate[Math Processing Error]?

Ans: In copper sulfate, we can see that there is one atom of copper so, we can say that 1 mole of CuSO4 will have 1 mole of copper.
The molar mass of copper sulphate is calculated below:
CuSO4 = 63.5 + 32 + (4 × 16)
= 63.5 + 32.0 + 64.0
= 159.5 g
We can say that 159.5 g of CuSO4 will have 63.5 g of copper.
⇒100 g of CuSO4 will contain 63.5 × 100 g159.5 of copper.
So, the amount of copper that can be obtained from 100 g of CuSO4=63.5 × 100 g159.5
= 39.81 g

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8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is

Ans: We are given the mass percentage of iron (Fe) as 69.9% and the percentage of oxygen (O) as 30.1%.
We know the atomic mass of iron is 55.85.
The number of moles of iron present in the oxide will be =69.9055.85
=1.25
The atomic mass of oxygen is 16.
The number of moles of oxygen present in the oxide will be =30.116.0
=1.88
The ratio of iron to the oxygen in the oxide is calculated below:
= 1.25: 1.88
=1.251.25: 1.881.25
= 1: 1.5
= 2: 3
According to the ratio the formula of the oxide will be Fe2O3.
Empirical formula mass of Fe2O3 = [ (2 × 55.85) + (3 × 16.00) ] = 159.7 g/ mol
Given molar mass ofFe2O3 = 159.69 g/ mol
Multiplying the empirical formula by n gives the molecular formula for the molecule. This is calculated below:
n = Molar massEmpirical formula mass=159.69 g159.7 g
= 0.999 = 1
So, the empirical formula of the oxide is Fe2O3 and the value of n is 1. Therefore, the molecular formula of the oxide is Fe2O3.

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9. Calculate the atomic mass (average) of chlorine using the following data:

Ans:The average atomic mass of chlorine is calculated below:
=[(Fractional abundance of 35Cl)(Molar mass of 35Cl)+(Fractional abundance of 37Cl)(Molar mass of 37Cl)]=[{(75.77100)(34.9689u)}+{(24.23100)(36.9659u)}]
= 26.4959 + 8.9568
= 35.4527 u
So, the average atomic mass of chlorine is 35.4527 u.

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10. In three moles of ethane calculate the following:
i). Number of moles of carbon atoms.

Ans: 2 Moles of carbon atoms are present in each mole of
Therefore, number of moles of carbon atoms in 3 moles of
ii) Number of moles of hydrogen atoms.
Ans: 6 moles of hydrogen atoms are present in each mole of [Math Processing Error]
Therefore, number of moles of carbon atoms in 3 moles of[Math Processing Error]
iii) Number of molecules of ethane.
Ans: [Math Processing Error] molecules of ethane are present in each mole of
Therefore, number of molecules in 3 moles of

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11. What is the concentration of sugar [Math Processing Error]in [Math Processing Error]if its 20 g are dissolved in enough water to make a final volume up to 2 L?<

Ans: Molarity (M) of the Ans can be calculated by the formula given below:
=Number of moles of soluteVolume of solution in Litres
This can be further written as
=Mass of sugar / molar mass of sugar2 L
Putting the values, we get:
=20 g/[(12 × 12)+(1 × 22)+(11 × 16)] g2 L
=20 g/ 342 g2 L
=0.0585mol2L
=0.02925 mol L-1
So, the molar concentration of sugar is =0.02925 mol L-1.

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12. If the density of methanol is 0.793 kg L-1, what is its volume needed for making 2.5 L of its 0.25 M Ans?

Ans: The molar mass of methanol is calculated below: (CH3OH)=(1×12)+(4×1)+(1×16)
=32 g mol-1
=0.032 kg mol-1
Molarity of the methanol will be =0.793 kg L-10.032 kg mol-1
=24.78 mol L−1
(Since density is mass per unit volume)
Now, to find the volume we have to write the formula,
M1V1=M2V2\
1 is for the given Ans and 2 is for the Ans to be prepared.
Putting the values, we get:
(24.78 mol L−1)V1=(2.5L)(0.25 mol L−1)
V1=0.0252 L
V1=25.22 mL

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13. Pressure is determined as force per unit area of surface. The SI unit of pressure,Pascal is as shown below:

Ans: Force per unit area of a surface is described as pressure. This is calculated below:
P=FA
=1034g × 9.8 ms-2cm2 × 1 kg1000 g × (100)2 cm21m2
=1.01332×105 kg m−1 s−2
We know the relation of force can be written as:
1 N=1 kg m s-2
Then we can write,
1 Pa=1 Nm-2=1 kg m-2 s-2
1 Pa=1 kg m-1 s-2
Pressure=1.01332 × 105 Pa

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14. What is the SI unit of mass? How is it defined?

Ans: The kilogram is the SI unit of mass in the SI system (kg). An international kilogram prototype is defined as one kilogram.

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15. Match the following prefixes with their multiples:

	Prefixes	Multiples
1.	Micro	106
2.	Deca	109
3.	Mega	44722
4.	Giga	42278
5.	Femto	10

Ans: The multiples are matched with their prefixes in the table given below:

	Prefixes	Multiples
1	Micro	10/06/22
2	Deca	10
3	Mega	106
4	Giga	109
5	Femto	01/10/15

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16. What do you mean by significant figures?

Ans: Significant figures are those digits that have meaning and are recognized to be assured of their value.
They are used to show uncertainty in an experiment or a computed number.
For example, if 18.2 mL is the result of an experiment, then 18 is certain while 2 is uncertain, and the total number of significant figures is 3.
Since the final digit signifies uncertainty, significant figures are defined as the sum of all of a number's decimal places.

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17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
i) Express this in percent by mass.

Ans: 1 part out of 1 million (106)parts is equal to 1 ppm.
Therefore, mass percent of 15 ppm chloroform in water will be
=15106 × 100
≃1.5×10−3
ii) Determine the molality of chloroform in the water sample.
Ans: 100 g of the sample will have 1.5 × 10−3of CHCl3.
So 1000 g of the sample contains will have 1.5 × 10−2of CHCl3 .
Therefore, molality of chloroform in water is calculated as:
=1.5 × 10-2 gMolar mass of CHCl3
For we want molar mass of chloroform.
Molar mass of CHCl3=12.00+1.00+3(35.5) = 119.5 g mol-1
∴ Molality of chloroform in water=0.0125 × 10-2m
=1.25 × 10-4m

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18. Express the following in the scientific notation:

Ans: If we want to express the numbers in scientific notation, we must first put the number in decimal form and multiply it by 10 with some power. These are given below:
0.0048
Ans: 0.0048=4.8×10−3
234,000
Ans: 234,000=2.34×105
8008
Ans: 8008=8.008×103
500.0
Ans: 500.0=5.000×102
6.0012 Ans: 6.0012=6.0012×100

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19. How many significant figures are present in the following?

Ans: There are some rules to find the number of significant figures and by following the rules the significant figures are given below:
1. 0.0025
Ans: There are 2 significant figures.
2. 208
Ans: There are 3 significant figures.
3. 5005
Ans: There are 4 significant figures.
4. 126,000
Ans: There are 3 significant figures.
5. 500.0
Ans: There are 4 significant figures.
6 .0034
Ans: There are 5 significant figures.

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21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
a) Which law of chemical combination is obeyed by the above experimental data? Give it a statement.

Ans: The masses of dioxygen that will combine with the fixed dinitrogen mass are 32 g, 64 g, 32 g, and 80 g, respectively. The mass ratios of dioxygen are 1: 2: 1: 5. In light of this fact, the supplied experimental data is in accordance with the law of multiple proportions. One element's mass multiplied by the fixed mass of another element must be small whole numbers if two elements combine to produce more than one compound, according to this rule.
b) Fill in the blanks in the following conversions:

Ans: All the conversions are given below:
i) 1 km = …………….. mm …………..pm
Ans: 1 km = 1 km × 1000 m1 km × 100 cm1 m × 10 mm1 cm
1km = 106mm
1 km = 1 km × 1000 m1 km × 1 pm10-12 m
1km=1015pm
Hence, 1 km = 106 mm = 1015 pm
ii) 1 mg = ……………... kg …………… ng
Ans: 1 mg = 1 mg × 1 g1000 mg × 1 kg1000 g
1mg = 106 kg
Hence, 1mg = 10-6 kg = 106 ng
iii) 1 mL = ……………. L ……………… dm3
Ans: 1 mL = 1 mL × 1 L1000 mL
1 mg = 10-3 L
1 mL = 1 cm3 = 1 cm3 × 1 dm × 1 dm × 1 dm10 cm × 10 cm × 10 cm
1 mL = 10-3 dm3
Hence, 1 mL = 10-3 L = 10-3 dm3

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22. If the speed of light is 3.0 × 108 ms-1, calculate the distance covered by light in 2.00 ns.

Ans: From the question we can see that the time taken to cover the distance is 2.00ns. This can be written as:
=2.00 × 10-9s
We know the speed of light= 3.0 × 108 ms-1
So, the distance travelled by light in 2.00ns will be:
=Speed of light × Time taken
=(3.0 × 108ms-1)(2.00 × 10-9s)
=6.00 × 10-1m
= 0.600 m
Identify the limiting reagent, if any in the following reaction mixtures.
Ans: Reactants that serve as limiting agents restrict the amount of a response. An initial reaction product is consumed before any more products are created, which stops the reaction and limits the quantity of products that are formed.
(i) 300 atoms of A + 200 molecules of B
Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. In other words, 200 molecules of B will react with 200 atoms of A, leaving 100 atoms of A unusable as a result. B is, thus, the limiting reagent in the reaction.
(ii) 2 mol A + 3 mol B
Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. In other words, 2 moles of A will react with just 2 moles of B. Consequently, 1 mol of A will not be used up in the process. This means that A is considered to be the limiting reagent.
(iii) 100 atoms of A + 100 molecules of B
Ans: According to the given reaction, 1 atom of A reacts with 1 molecule of B. The mixture is stoichiometric where no limiting reagent is present.
(vi) 5 mol A + 2.5 mol B
Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. Because of this, only 2.5 mol of B may be combined with 2.5 mol of A. There will be 2.5 mol of a remaining. B is, thus, the limiting reagent in this reaction.
(v). 2.5 mol A + 5 mol B
Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. Because of this, only 2.5 mol of A may be combined with 2.5 mol of B. There will be 2.5 mol of a remaining. A is, thus, the limiting reagent in this reaction

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24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2(g)+H2(g)→2NH3(g)
(i). Calculate the mass of ammonia produced if 2.00 × 103g dinitrogen reacts with 1.00 × 103gof dihydrogen

Ans: First, we have to balance the given chemical equation.
N2(g)+3H2(g)→2NH3(g)
From this equation we can say, 1 mole (28g) of dinitrogen reacts with 3 mole (6g) of dihydrogen to give 2 mole (34g)of ammonia.
=2.00 ×103 gof nitrogen will react with6 g28 g × 2.00 × 103 g dihydrogen.
=2.00 × 103g of dinitrogen will react with 428.6 g of dihydrogen.
We are given the amount of dihydrogen = 1.00 × 103 g
Therefore,N2is the limiting reagent.
We can say that 28 g of N2 will produce 34 g of NH3.
Hence, mass of ammonia produced by 2000 g of N2 =34 g28 g × 2000 g = 2428.57 g
(i) Will any of the two reactants remain unreacted?
Ans: N2 is the limiting reagent whileH2 is the excess reagent. So, hydrogen will be left unreacted.
(ii) If yes, which one and what would be its mass?
Ans: Mass of dihydrogen left unreacted can be calculated as
=1.00 × 103g - 428.6g
= 571.4 g

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25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3different?

Ans: The molar mass of Na2CO3 is given below:
Na2CO3 = (2 × 23) + 12.00 + (3 × 16)
= 106 g/mol
So, 1 mole of Na2CO3 means 106 g of Na2CO3.
Therefore, for 0.5 mol of Na2CO3 can be calculated as
0.5 mol of Na2CO3 = 106 g1 mole × 0.5 mol Na2CO3
=53g Na2CO3
= 0.50 M Na2CO3=0.50mol/L Na2CO3
Hence,0.50 mol Na2CO3is present in 1 L of water or 53g Na2CO3is present in 1 L of water.

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26. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapor would be produced?

Ans: Let us first write the reaction between dihydrogen and dioxygen. The reaction will be:
2H2(g)+O2(g)→2H2O(g)
Dioxygen reacts with two volumes of dihydrogen to generate two volumes of water vapour.
As a result, ten volumes of dihydrogen will react with five volumes of dioxygen to generate ten volumes of water Vapours.

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27. Convert the following into basic units:

Ans:To convert the given numbers into basic units we have to convert it into meters and kilograms. All the numbers are converted into meters and kilograms, and are written below:
(i) 28.7 pm
Ans: 1pm=10-12m
28.7pm=28.7 × 10-12m
=2.87 × 10-11m
(ii) 15.15 pm
Ans: 1pm=10-12m
15.15pm=15.15 × 10-12m
=1.515 × 10-12m
(iii) 25365 mg
Ans: 1mg = 10-3g
25365mg=2.53657 × 104 × 10-3g
Since,
1g=10-3kg
2.5365 × 101g=2.5365 × 10-1 × 10-3kg
25365mg=2.5365 × 10-2kg$$

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28. Which one of the following will have the largest number of atoms?

Ans: As we can see that all the atoms are given in grams so, by counting the number of atoms in each element we can find the answer. These are solved step by step and written below:
(i) 1 g Au (s)
Ans: 1g of Au (s)=1197mol of Au (s)
=6.022 × 1023197 atoms of Au (s)
=3.06 × 1021 atoms of Au (s)
(ii) 1 g Na (s)
Ans: 1g of Na (s)=123mol of Na (s)
=6.022 × 102323 atoms of Na (s)
=0.262 × 1023 atoms of Na (s)
=26.2 × 1021 atoms of Na (s)
(iii) 1 g Li (s)
Ans: 1g of Li (s)=17mol of Li (s)
=6.022 × 10237 atoms of Li (s)
=0.86 × 1023 atoms of Li (s)
=86.0 × 1021 atoms of Li (s)
(vi) 1 g of Cl2 (g)
Ans: 1g of Cl2 (g)=171mol of Cl2 (g)
Molar mass of Cl2 is 71 (2 × 35.5)
=6.022 × 102371 atoms of Cl2 (g)
=0.0848 × 1023 atoms of Cl2 (g)
=8.48 × 1021 atoms of Cl2 (g)
Hence, 1 g of Li (s) has the largest number of atoms.

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29. Calculate the molarity of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Ans: First we have to find the mole fraction of ethanol.
Mole fraction of C2H5OH=Number of moles C2H5OHNumber of moles of solution
0.40=nC2H5OHnC2H5OH+nH2O .................(1)
Number of moles present in 1 L water:
nH2O=1000 g18 g mol-1
nH2O=55.55mol
Substituting the value of nH2Oin equation (1)
nC2H5OHnC2H5OH+55.55=0.040
nC2H5OH=0.040nC2H5OH+(0.040)(55.55)
0.96nC2H5OH=2.222mole
nC2H5OH=2.314mole
Molarity of solution =2.314 mol1 L So, the molarity of the Ans is 2.314 M.

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30. What will be the mass of one 12C atom in g?

Ans: We know that 1 mole of carbon means 12 gram of carbon is there and in terms of number of atoms, there are 6.023 × 1023 atoms in 1mole of carbon. This can be written as:
\ 1 mole = 12 g = 6.023 × 1023
So, mass of one12C atom=12g6.022 × 1023
=1.993 × 10-23g

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31. How many significant figures should be present in the answer of the following calculations?

Ans: First we have to find the least precise number to find the significant figures.
(i) 0.2856 × 298.15 × 0.1120.5785
Ans: 0.2856 × 298.15 × 0.1120.5785
Least precise number of calculation = 0.112
Therefore, number of significant figures in the answer
Number of significant figures in the least precise number = 3
(ii) 5 × 5.364
Ans: 5 × 5.364
Least precise number of calculations = 5.364
Therefore, number of significant figures in the answer will be
Number of significant figures in 5.364 = 4
(iii) 0.0125+0.7864+0.0215
Ans: 0.0125+0.7864+0.0215
Since the least number of decimal places in each term is four, the number of significant figures in the answer will also be 4.

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32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Isotope	Isotopic molar mass	Abundance
36Ar	35.96755g mol-1	0.00337
38Ar	37.96272 g mol-1	0.00063
40Ar	39.9624 g mol-1	0.996

Ans: Molar mass of argon is calculated step by step.
=[(35.96755 × 0.337100)+(37.96272 × 0.063100)+(39.9624 × 90.60100)]g mol-1
=[0.121+0.024+39.802]g mol-1
So, the molar mass of argon is 39.947 g/ mol.

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33. Calculate the number of atoms in each of the following

Ans: All the options are given in different forms so, we have to convert it into a number of atoms. These are given below:
(i) 52 moles of Ar
Ans: 1 mole of Ar = 6.022 × 1023 atoms of Ar
Therefore,
52 mole of Ar = 52 × 6.022 × 1023 atoms of Ar
=3.131 × 1025 atoms of Ar
(ii) 52 u of He
Ans: 1 atom of He = 4u of the He
Or this can be written as, 4 u of He = 1 atom of He
1. u of He = 14 atom of He
of He = 524 atom of He
= 13 atoms of He.
(iii) 52 g of He
Ans: 4g of He = 6.022 × 1023 atoms of He
Therefore, we can write:
52g of He = 6.022 × 1023 × 524 atoms of He
= 7.8286 × 1024 atoms of He

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34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide 0.690 g water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

(i) Empirical formula
Ans: 1 mole (44 g) of CO2 will have 12 g carbon.
So, 3.38 g of CO2 will have carbon = 12g44g × 3.38g
= 0.9217 g
18 g of water will have 2 g of hydrogen.
So, 0.690 g of water contain hydrogen = 2g18g × 0.690
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:
= 0.9217 g + 0.0767 g
=0.9984 g
So, the percentage of Carbon in the compound = 0.92170.9984 × 100= 92.32%
Now, percentage of Hydrogen in the compound = 0.07670.9984 × 100= 7.68%
Moles of carbon in the compound = 92.3212=7.69
Moles of hydrogen in the compound = 7.681=7.68
Since, we have the number of moles of both the elements, the ratio of carbon to hydrogen will be:
7.69: 7.68 = 1: 1
This is in the whole number, so the empirical formula will be CH.
(ii) Molar mass of the gas
Ans: We are given,
Weight of 10.0 L of the gas (at S.T.P) = 11.6 g
Weight of 22.4L of gas at STP = 11.6g10.0L × 22.4L
=25.984g
≈26g
Hence, the molar mass of the gas is 26 g.
(iii) Molecular formula.
Ans: Empirical formula mass of CH = 12 + 1 = 13
Now, we can find the value of n. This is given below:
n = Molar mass of gasEmpirical formula mass of gas
=26g13g
n = 2
Therefore,
Molecular formula of gas=(CH)n
= C2H2

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35. Calcium carbonate reacts with aqueous HCl to givenCaCl2 and CO2 according to the reaction,
CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Ans: We are given the molarity as 0.75 M which means in 1 L of Ans there are 0.75 moles of HCl. This can be converted into grams as:
=[(0.75 mol) × (36.5gmol-1)]HCl is present in 1L of water
=27.375g of HCl is present in 1L of water
Thus, 1000 mL of Ans contains 27.375 g of HCl.
So, the amount of HCl present in 25 mL of Ans
=27.375g1000mL × 25mL
= 0.6844 g
The chemical reaction given in the question is:
CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)
2 mol of HCl (2 × 36.5=73g)reacts with 1 mol of CaCO3(100g) .
So, the amount ofCaCO3 that will react with 0.6844g = 10071 × 0.6844g = 0.9639 g

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36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction
4HCl(aq)+MnO2(l)→MnCl2(aq)+Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?

Ans: 1 mol of MnO2 reacts with 4 mol of HCl.
The molar mass of MnO2is 87g and the molar mass of HCl is 36.5 g. For 4 moles of HCl, the mass will be 146 g.
So, for 5.0 g of MnO2 will react with:
=14687 × 5 = 8.4 g
Therefore, 8.4 g of HCl will completely react with 5 g of MnO2.

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Last Updated on: Mar 28, 2024