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  4. The degree of dissociation of Ca(NO3)2 in dilute aqueous solution containing 7.0 g of salt per 100.0 g of water at 100°C is 70 %. If the vapour pressure of water at 100°C is 760 mmHg, the vapour pressure of the solution is

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Q. The degree of dissociation of $Ca(NO_{3})_{2}$ in dilute aqueous solution containing 7.0 g of salt per 100.0 g of water at $100^{\circ}C$ is $70\%$. If the vapour pressure of water at $100^{\circ}C$ is 760 mmHg, the vapour pressure of the solution is

Solutions

Solution:

Moles of $Ca(NO_{3})_{2}=\frac{7}{168}$ present in 100.0 g of water
Since dissociation is 70%, the total number of particles
$\frac{7}{168}\times 0.7 \times 3 = 0.0875=n$
Also, moles of solvent $=\frac{100}{18}=5.55=N$
Applying Raoult’s law
$\frac{p^{\circ}-p_{\text{solution}}}{P^{\circ}}=\frac{n}{n+N}$
Or $\frac{760-p}{760}=\frac{0.087}{0.087+5.55}$
Or $760-P-760\left[\frac{0.087}{5.637}\right]=760\left[0.0154\right]=11.704$
Or $P=760 - 11.704=748.2\,mmHg$