Verified Solution • 2025–26 Session moderate

QJEEM22Ch1562: JEE Main 2022 Chemistry - Class 12 JEEM22Ch1562 Solution: A solution containing 2.5 10 3 kg of a solute dissolved in 75…

A solution containing 2.5 ×\times 10 −- 3 kg of a solute dissolved in 75 ×\times 10 −- 3 kg of water boils at 373.535 K. The molar mass of the solute is ____________ g mol −- 1. [nearest integer] (Given : Kb(H2O) = 0.52 K kg mol −- 1 and boiling point of water = 373.15 K)

Step-by-Step Verified Solution

Wsolute =2.5×10−3 kgW_{\text {solute }}=2.5 \times 10^{-3} \mathrm{~kg}

amp;Wsolvent =75×10−3 kgamp;ΔTb=373.535−373.15amp;=0.385 Kamp; Kb(H2O)=0.52 K kg mol−1amp;ΔTb=Kb×103×Wsolute Msolute ×Wsolvent amp;Msolute =0.52×103×2.5×10−375×10−3×0.385amp;=45.02amp;≈45\begin{aligned}&\mathrm{W}_{\text {solvent }}=75 \times 10^{-3} \mathrm{~kg} \\\\&\Delta \mathrm{T}_{\mathrm{b}} =373.535-373.15 \\\\&=0.385 \mathrm{~K} \\\\&\mathrm{~K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right) =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\\\&\Delta \mathrm{T}_{\mathrm{b}} =\frac{\mathrm{K}_{\mathrm{b}} \times 10^{3} \times \mathrm{W}_{\text {solute }}}{\mathrm{M}_{\text {solute }} \times \mathrm{W}_{\text {solvent }}} \\\\&\mathrm{M}_{\text {solute }} =\frac{0.52 \times 10^{3} \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385} \\\\&=45.02 \\\\& \approx 45\end{aligned}

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