Question

What is the freezing point of a solution that contains $\text{10} \text{.0} \, \text{g}$ of a glucose $\left( C _6 H _{12} O _6\right)$ in $100 g$ of $H _2 O ? K _{ f }=1.86^{\circ} C / m$

• A. $-0.186^\circ C$
• B. $+0.186^\circ C$
• C. $-0.10^\circ C$
• D. $-1.03^\circ C$

Answer 1

Answer: (D)

$\Delta T_{f}=K_{f}m=K_{f}\times \frac{W_{B}}{M_{B} . W_{A}}$
$=\frac{1.86 \text{ K kg mo} \left(\text{l}\right)^{- 1} \times 10.0 \text{ g}}{180 \text{ g mo} \left(\text{l}\right)^{- 1} \times 100 \text{ g} \times \left(\frac{1 \text{ kg}}{1000 \text{ g}}\right)}$
$=\frac{1.86 \times 10.0 \times 1000 \text{ K}}{180 \times 100}$
$=1.03\text{ K}=1.03^\circ C$
F.P. of solution $=F.P$ of water $-\Delta T_{f}$
$=0^\circ C-1.03^\circ C=-1.03^\circ C$