Question

A very small mass of a non-volatile solute (that does not dissociate) is dissolved in $56.8c{m}^3$ of benzene (density is $0.889gc{m}^{-3})$. At room temperature, vapour pressure of this solution is 98.88 mm Hg, while that of benzene is 100 mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal depression constant of benzene?

• A. $7Kmolalit{y}^{-1}$
• B. $5Kmolalit{y}^{-1}$
• C. $3.5Kmolalit{y}^{-1}$
• D. none of these

Answer 1

The correct option is B $5Kmolalit{y}^{-1}$

As we know,

$\frac{P^0-P_S}{P_S}=\frac{w\times M}{m\times W}$

$\therefore \frac{100-98.88}{98.88}=\frac{w\times 78\times 1000}{m\times W\times 1000}$

or $molality=\left(\frac{w\times 1000}{m\times W}\right)$

$\frac{P^o-P_S}{P_S}\times \frac{1000}{M}=\frac{1.12\times 1000}{78\times 98.88}=0.1452$

Also, $\Delta T=K_f\times molality$

$0.73=K_f\times 0.1452$

$\therefore K_f=5.027Kmolalit{y}^{-1}$