The degree of dissociation of Ca(NO_ 3 )_ 2 in dilute aq. solution containing 7.0 g of salt per 100 g of water at 100^ ^ C is 70 \% . If vapour pressure of water at 100^ ^ C is 760 mm Hg . The vapour pressure of

Correct answer: D. 746 Ca(NO_ 3 )_ 2 leftharpoons Ca^ 2+ +2 NO_ 3 ^ - 1-0.71 0.7 2 × 0.7 ( α=(70)/(100)=0.7) n_ 1 =(100)/(18)=5.55 ^ 0 -p_ s ^ 0 = n_ 2 × i n_ 1 +n_ 2 , 760-p_ s 760 =(0.042 × 2.4)/(5.55+0.042) p_ s -746 mm Hg.

BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Chemistry
Chapter: Solutions

BITSAT 2009 · Chemistry · Solutions

The degree of dissociation of Ca(NO_ 3 )_ 2 in dilute aq. solution...

The degree of dissociation of

\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}

in dilute aq. solution containing

7.0 \mathrm{ g}

of salt per

100 \mathrm{ g}

of water at

100^{{}^{\circ}} \mathrm{C}

70 \%

. If vapour pressure of water at

100^{{}^{\circ}} \mathrm{C}

760 \mathrm{ mm} \mathrm{Hg}

. The vapour pressure of solution is

Options

A735
B $\quad 730$
C760
D $\quad 746$

Correct answer

D. $\quad 746$

Step-by-step solution

\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{NO}_{3}^{-}1-0.71 \quad 0.7 \quad 2 \times 0.7\quad\left(\because \alpha=\frac{70}{100}=0.7\right)\quad \mathrm{n}_{1}=\frac{100}{18}=5.55\frac{\mathrmp^{0}-\mathrm{p}_{\mathrm{s}}}{\mathrmp^{0}}=\frac{\mathrm{n}_{2} \times \mathrm{i}}{\mathrm{n}_{1}+\mathrm{n}_{2}}, \frac{760-\mathrm{p}_{\mathrm{s}}}{760}=\frac{0.042 \times 2.4}{5.55+0.042}\therefore \mathrm{p}_{\mathrm{s}}-746 \mathrm{ mm} \mathrm{Hg}.

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