Calculate the mass of CaCi, (molar mass = 111 g mol-1) to be dissolved…

Calculate the mass of CaCi, (molar mass = 111 g mol-1) to be dissolved in 500 g of water to lower its freezing point by 2K, assuming that CaCl, undergoes complete dissociation.
(Kf for water = 1.86 K kg mol-1)

Ans.

To calculate the mass of calcium chloride (����2CaCl2​) needed to lower the freezing point of 500 g of water by 2 K, we can use the formula for freezing point depression:

Δ��=�⋅��⋅�ΔTf​=i⋅Kf​⋅m

Where:

• Δ��ΔTf​ is the freezing point depression,

• $i$ is the van’t Hoff factor,

• ��Kf​ is the freezing point depression constant for water, and

• $m$ is the molality of the solution.

Given:

• Δ��=2ΔTf​=2 K

• ��=1.86Kf​=1.86 K kg mol−1−1 (we need to convert this to molal units)

• $m$ is the molality of the solution

First, let’s calculate the molality ($m$) of the solution:

Δ��=�⋅��⋅�ΔTf​=i⋅Kf​⋅m

Rearranging the equation to solve for $m$:

�=Δ���⋅��m=i⋅Kf​ΔTf​​

We know that $i$ for ����2CaCl2​ is 3 because it dissociates into three ions (1 calcium ion and 2 chloride ions).

So,

�=2 K3×1.86 K kg mol−1m=3×1.86K kg mol−12K​

�=25.58≈0.358 mol kg−1m=5.582​≈0.358mol kg−1

Now, we can calculate the number of moles of ����2CaCl2​ needed:

Molality=moles of solutemass of solvent in kgMolality=mass of solvent in kgmoles of solute​

0.358 mol kg−1=moles of ����20.500 kg0.358mol kg−1=0.500kgmoles of CaCl2​​

moles of ����2=0.358×0.500 mol≈0.179 molmoles of CaCl2​=0.358×0.500mol≈0.179mol

Now, we can calculate the mass of ����2CaCl2​ using its molar mass:

$$\text{mass}=\text{moles}\times \text{molar mass}$$

mass=0.179×111 g mol−1≈19.869 gmass=0.179×111gmol−1≈19.869g

So, approximately 19.869 grams of ����2CaCl2​ need to be dissolved in 500 grams of water to lower its freezing point by 2 K.