The vapour pressure of 30 \%(w / v) aqueous solution of glucose is \underline\hspace2cm mm ~Hg at 25^\circ C . [Given : The density of 30 \% (w/v), aqueous solution of glucose is 1.2 ~g ~cm^-3 and vapour pressure of pure water is 24 ~mm~ Hg .] (Molar mass of glucose is 180 ~g ~mol^-1 .)

Solutions PYQ — JEE Main 2023 · 15 Apr · Shift 1 · Q19 with Solution

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Correct answer: 23

A $30\%\,(w/v)$ glucose solution means:

$30\text{ g glucose in }100\text{ mL solution}$

Given density of solution $=1.2\text{ g cm}^{-3}=1.2\text{ g mL}^{-1}$ .

So, mass of $100\text{ mL}$ solution is

$100\times 1.2=120\text{ g}$

Out of total $120\text{ g}$ solution, glucose is $30\text{ g}$ .

Hence mass of water is

$120-30=90\text{ g}$

$n_{\text{glucose}}=\frac{30}{180}=\frac{1}{6}\text{ mol}$

$n_{\text{water}}=\frac{90}{18}=5\text{ mol}$

For a non-volatile solute:

$P_{\text{solution}}=X_{\text{water}}\,P^0_{\text{water}}$

where

=\frac{5}{5+\frac{1}{6}}$$ $$X_{\text{water}}=\frac{5}{\frac{31}{6}}=\frac{30}{31}$$ So, $$P_{\text{solution}}=\frac{30}{31}\times 24$$ $$P_{\text{solution}}=23.23\text{ mm Hg}$$ 6. **Final integer answer** $$\boxed{23}$$ Thus, the vapour pressure of the solution is **23 mm Hg**.