18 g of a non-volatile solute is dissolved in 200 g of H, freezes at 272.07…

18 g of a non-volatile solute is dissolved in 200 g of H, freezes at 272.07 K. Calculate the molecular mass of solute (K, for water = 1.86 K kg mol-1)

Ans.

To solve this problem, we can use the concept of freezing point depression.

The freezing point depression (ΔTf​) is given by the formula:

ΔTf​=Kf​⋅m

where:

• ΔTf​ is the freezing point depression,

• Kf​ is the cryoscopic constant (which is 1.86 K kg mol−11.86K kg mol−1 for water),

• m is the molality of the solution.

We first need to find the molality (m) of the solution, which is the number of moles of solute per kilogram of solvent.

Given:

• Mass of solvent(water), mH2​O​=200g=0.2kg

• Mass of solute, msolute​=18g

Molar mass of solvent (water), MH2​O​=18g/mol

Moles of solute, nsolute​=Msolute​msolute​​​​

Molality, m=msolvent​nsolute​​​​

Now, we can calculate the freezing point depression (ΔTf​​) using the given cryoscopic constant and the molality of the solution.

Finally, the freezing point of the solvent is decreased by ΔTf​​ from its normal freezing point of $273.15\text{K}$ to 273.15−Δ��273.15−ΔTf​ K.

Let’s calculate:

First, find the molality: �solute=18 g�solutensolute​=Msolute​18g​ �solute=18 g18 g/mol=1 molnsolute​=18g/mol18g​=1mol

�=�solute�solvent=1 mol0.2 kg=5 mol/kgm=msolvent​nsolute​​=0.2kg1mol​=5mol/kg

Now, calculate the freezing point depression: Δ��=��⋅�=1.86 K kg mol−1×5 mol/kg=9.3 KΔTf​=Kf​⋅m=1.86K kg mol−1×5mol/kg=9.3K

The freezing point of the solvent decreases by $9.3\text{K}$ from its normal freezing point of $273.15\text{K}$ to $273.15-9.3=263.85\text{K}$.

Therefore, the molecular mass of the solute is $18\text{g/mol}$.