Chemistry - Liquid Solution Question with Solution | TestHub

To calculate the vapour pressure of the $30\%$ (w/v) aqueous solution of glucose, we can use Raoult's law, which states that the vapour pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.

Weight of solution $=100\times 1.2=120 \mathrm{gm}$

Weight of water $=120-30=90 \mathrm{gm}$

Now using formula

$\frac{{\mathrm{P}}_0-\mathrm{P}}{\mathrm{P}}=\frac{\mathrm{moles} \mathrm{of} glucose}{\mathrm{moles} \mathrm{of} water}$

$\frac{24-\mathrm{P}}{\mathrm{P}}=\frac{\frac{30}{180}}{\frac{90}{18}}=\frac{3}{90}$

$24\times 90-\mathrm{P}\times 90=3 \mathrm{P}$

$\Rightarrow \mathrm{P}=23.22$