0.5 molal aqueous solution of a weak acid (HX) is 20 \% ionised. If K_f for water is 1.86 Kg mol^ -1 , the lowering in freezing point of the solution is:

Correct answer: B. 1.12 ~K As Δ T_f=i k_f m For aligned & HX leftharpoons H^ + +X^ - & t=0 1 0 0 & array llll t_ eq & (1-0.20) & 0.20 & 0.20 array & aligned Total no. of moles =1-0.20+0.20 aligned +0.20 & =1+0.20=1.2 Δ T_f & =1.2 × 1.86 × 0.5 & =1.1160 aligned Related Theory The boiling point and freezing point are colligative properties. Due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.

NEET 2007

Exam: NEET
Year: 2007
Subject: Chemistry
Chapter: Solutions

NEET 2007 · Chemistry · Solutions

0.5 molal aqueous solution of a weak acid (HX) is 20 \% ionised. If...

0.5 molal aqueous solution of a weak acid

(\mathrm{HX})

20 \%

ionised. If

\mathrm{K}_f

for water is

1.86 \mathrm{Kg} \mathrm{mol}^{-1}

, the lowering in freezing point of the solution is:

Options

A $0.56 \mathrm{~K}$
B $1.12 \mathrm{~K}$
C $-0.56 \mathrm{~K}$
D $-1.12 \mathrm{~K}$

Correct answer

B. $1.12 \mathrm{~K}$

Step-by-step solution

\operatorname{As} \Delta \mathrm{T}_f=i k_f \mathrm{ m}

For

\begin{aligned} & \mathrm{HX} \rightleftharpoons \mathrm{H}^{+}+\mathrm{X}^{-} \\ & t=0 \quad 1 \quad 0 \quad 0 \\ & \begin{array}{llll} t_{\text{eq }} & (1-0.20) & 0.20 & 0.20 \end{array} \\ & \end{aligned}

Total no. of moles

=1-0.20+0.20

\begin{aligned} +0.20 & =1+0.20=1.2 \\ \therefore \Delta \mathrm{T}_f & =1.2 \times 1.86 \times 0.5 \\ & =1.1160 \end{aligned}

Related Theory The boiling point and freezing point are colligative properties. Due to the addition of solutes, the boiling point tends to increase, and freezing point tends to decrease.

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