Question

The difference between the boiling point and freezing point of an aqueous solution containing sucrose (molecular mass = $342g$ $mo{l}^{-1}$) in $100g$ of water is $105.0^{\circ }C$. If $K_{f}and{K}_b$ of water are $1.86$ and $0.51Kkgmo{l}^{-1}$ respectively, the weight of sucrose in the solution is about

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

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 , $T_b-T_f=(T_b^{\circ }+\Delta T_b)-(T_f^{\circ }-\Delta T_f)$
105 = $(\Delta T_b+\Delta T_f)+(T_b^{\circ }-T_f^{\circ })$
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 = $\frac{Molesofsolute}{Massofsolvent(kg)}$
$\therefore$ Moles of solute = 2.11 $\times$ 0.1 = 0.211
$\therefore$ Mass of solute = 0.211 $\times$ 342 = 72.16 g.