Question

"An aqueous CaCl2 solution has a vapor pressure of 83.1mmHg at 50 ∘C. The vapor pressure of pure water at this temperature is 92.6 mmHg. What is the concentration of CaCl2 in mass percent?"

Answer 1

Answer : The the concentration of $CaCl_2$ in mass percent is, 41.18 %

Solution : Given,

Molar mass of water = 18 g/mole

Molar mass of $CaCl_2$ = 110.98 g/mole

First we have to calculate the mole fraction of solute.

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

$\fac{p^o-p_s}{p^o}=X_B$

where,

$p^o$ = vapor pressure of the pure component (water) = 92.6 mmHg

$p_s$ = vapor pressure of the solution = 83.1 mmHg

$X_B$ = mole fraction of solute, $(CaCl_2)$

Now put all the given values in this formula, we get the mole fraction of solute.

$\fac{92.6-83.1}{92.6}=X_B$

$X_B=0.102$

Now we have to calculate the mole fraction of solvent (water).

As we know that,

$X_A+X_B=1\\\\X_A=1-X_B\\\\X_A=1-0.102\\\\X_A=0.898$

The number of moles of solute and solvent will be, 0.102 and 0.898 moles respectively.

Now we have to calculate the mass of solute, $(CaCl_2)$ and solvent, $(H_2O)$.

$\text{Mass of }CaCl_2=\text{Moles of }CaCl_2\times \text{Molar mass of }CaCl_2$

$\text{Mass of }CaCl_2=(0.102mole)\times (110.98g/mole)=11.32g$

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.898mole)\times (18g/mole)=16.164g$

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 11.32 + 16.164 = 27.484 g

Now we have to calculate the mass percent of $CaCl_2$

$Mass\%=\frac{\text{Mass of}CaCl_2}{\text{Mass of solution}}\times 100=\frac{11.32g}{27.484g}\times 100=41.18\%$

Therefore, the the concentration of $CaCl_2$ in mass percent is, 41.18 %