Q. A solution containing 2.5 g of a non-volatile solute in 100 g of water freezes at -0.465°C. The molecular weight of the solute is: (Kf for water = 1.86°C/m)

100 ग्राम पानी में 2.5 ग्राम गैर-वाष्पशील विलेय वाला एक घोल -0.465°C पर जम जाता है। विलेय का आणविक भार है: (पानी के लिए Kf = 1.86°C/m)

50 g/mol
50 ग्राम/मोल

100 g/mol
100 ग्राम/मोल

150 g/mol
150 ग्राम/मोल

200 g/mol
200 ग्राम/मोल

Calculate molality $m$ :
$\Delta T_f = K_f \times m \implies m = \frac{\Delta T_f}{K_f} = \frac{0.465}{1.86} = 0.25 \, \text{mol/kg}$
Moles of solute = molality × kg solvent = $0.25 \times 0.1 = 0.025$ mol
Molecular weight:
$M = \frac{\text{mass}}{\text{moles}} = \frac{2.5}{0.025} = 100 \, \text{g/mol}$

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