Medium INTEGER +4 / -1 PYQ · JEE Mains 2023

Sea water contains $29.25 \% ~\mathrm{NaCl}$ and $19 \% ~\mathrm{MgCl}_{2}$ by weight of solution. The normal boiling point of the sea water is _____________ ${ }^{\circ} \mathrm{C}$ (Nearest integer)

Assume $100 \%$ ionization for both $\mathrm{NaCl}$ and $\mathrm{MgCl}_{2}$

Given : $$\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$

Molar mass of $\mathrm{NaCl}$ and $\mathrm{MgCl}_{2}$ is 58.5 and 95 $\mathrm{g} \mathrm{~mol}^{-1}$ respectively.

Answer (integer) 116

Solution

Amount of solvent = 100 - (29.25 + 19) = 51.75 g <br/><br/> Now, we can calculate the boiling point elevation using the given formula: <br/><br/> ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52<br/><br/> ΔTb = 16.075 <br/><br/> The boiling point of the sea water is the normal boiling point of water plus the change in boiling point: <br/><br/> Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C <br/><br/> Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.

About this question

Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws

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