Chemistry - Redox Reactions Question with Solution | TestHub

Relative lowering of vapour pressure is given as
$\frac{{\mathrm{p}}^{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^{\mathrm{o}}}=\frac{{\mathrm{w}}_{\mathrm{A}}}{{\mathrm{m}}_{\mathrm{A}}}\times \frac{{\mathrm{m}}_{\mathrm{B}}}{{\mathrm{w}}_{\mathrm{B}}}$
As relative lowering of vapour pressure
$=\frac{{\mathrm{p}}^{\mathrm{o}}-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^{\mathrm{o}}}=\frac{10}{100}$
${\mathrm{m}}_{\mathrm{A}}=60,{\mathrm{w}}_{\mathrm{B}}=180,{\mathrm{w}}_{\mathrm{A}}=\mathrm{x}$
So$\frac{10}{100}=\frac{\mathrm{x}/60}{180/18}$
$\frac{1}{10}=\frac{\mathrm{x}/60}{10}$
It mean$60$$\mathrm{g}$of the solute must be added to$180$$\mathrm{g}$of water so that the vapour pressure of water is lowered by 10 %.