Question

The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular mass = $342 g$ $mol^{-1}$) in $100\,g$ of water is $105 .0^\circ C$. If $K_f \,and\, K_b$ of water are $1.86$ and $0.51 \,K \,kg mol^{-1}$ respectively, the weight of sucrose in the solution is about

• A. $34.2\, g$
• B. $342\, g$
• C. $7.2\, g$
• D. $72\, g$
• E. $68.4\, g$

Answer 1

Answer: (D)

$\Delta \, T_b = K_b \times m$ and $\Delta \, T_f = K_f \times m$
$\therefore $ $\Delta \, T_b + T_f = (K_b +K_f)m$
Now , $T_b - T_f = (T^\circ _b + \Delta \, T_b) - (T^\circ _f - \Delta \,T_f)$
105 = $(\Delta \, T_b +\Delta \, T_f) + (T^\circ _b - T^\circ _f)$
105 = $(\Delta \, T_b +\Delta \, T_f) + 100$
$\therefore $ $\Delta \, T_b + \Delta \, T_f = 5 $
$\therefore $ $m = \frac{\Delta T_b + \Delta_f}{K_b + K_f} = \frac{5}{1.86 + 0.51} = \frac{5}{2.37} = 2.11$
Molality = $\frac{Moles \, of \,solute}{Mass \, of \, solvent (kg)}$
$\therefore $ Moles of solute = 2.11 $\times $ 0.1 = 0.211
$\therefore $ Mass of solute = 0.211 $\times$ 342 = 72.16 g.