Q. The freezing point (in$℃$) of solution containing 0.1 g of $K_{3}\left\lbrack Fe(CN)_{6} \right\rbrack(mol.wt\ 329)$ in 100 g of water ($K_{f} = 1.86\ K\ kg\ mol^{- 1}$)is
A) $- 2.3 \times 10^{- 2}$
B) $- 5.7 \times 10^{- 2}$
C) $- 5.7 \times 10^{- 3}$
D) $- 1.2 \times 10^{- 2}$
Explanation
(a)
Van’t Hoff’s factor (i)=4 {$3K^{+}\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -}\}$
Molality =$\frac{0.1}{329} \times \frac{1000}{100} = \frac{1}{329}$
$\Longrightarrow$ -$\bigtriangleup T_{f} = iK_{f}.m$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4 \times 1.86 \times \frac{1}{329} = 2.3 \times 10^{- 2}$
$\Longrightarrow$ $\ \ \ \ \ \ T_{f} = - 2.3 \times 10^{- 2}℃$
(As freezing point of water is $0^{\circ}C$)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.