JEE MAIN - Chemistry (2022 - 27th June Evening Shift - No. 18)
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)
Answer
45
Explanation
$W_{\text {solute }}=2.5 \times 10^{-3} \mathrm{~kg}$
$$ \begin{aligned} &\mathrm{W}_{\text {solvent }}=75 \times 10^{-3} \mathrm{~kg} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =373.535-373.15 \\\\ &=0.385 \mathrm{~K} \\\\ &\mathrm{~K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right) =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =\frac{\mathrm{K}_{\mathrm{b}} \times 10^{3} \times \mathrm{W}_{\text {solute }}}{\mathrm{M}_{\text {solute }} \times \mathrm{W}_{\text {solvent }}} \\\\ &\mathrm{M}_{\text {solute }} =\frac{0.52 \times 10^{3} \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385} \\\\ &=45.02 \\\\ & \approx 45 \end{aligned} $$
$$ \begin{aligned} &\mathrm{W}_{\text {solvent }}=75 \times 10^{-3} \mathrm{~kg} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =373.535-373.15 \\\\ &=0.385 \mathrm{~K} \\\\ &\mathrm{~K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right) =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =\frac{\mathrm{K}_{\mathrm{b}} \times 10^{3} \times \mathrm{W}_{\text {solute }}}{\mathrm{M}_{\text {solute }} \times \mathrm{W}_{\text {solvent }}} \\\\ &\mathrm{M}_{\text {solute }} =\frac{0.52 \times 10^{3} \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385} \\\\ &=45.02 \\\\ & \approx 45 \end{aligned} $$
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