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The freezing point depression constant of water is Kf=1.86 K⋅kg/molK_f = 1.86\,\text{K}\cdot\text{kg/mol}. ΔTf\Delta T_f for 0.1 m0.1\,\text{m} NaClNaCl (i=2i=2) is:

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Answer: (B) 0.372 K0.372\,\text{K}

ΔTf=iKfm=2×1.86×0.1=0.372 K\Delta T_f = i K_f m = 2 \times 1.86 \times 0.1 = 0.372\,\text{K}.

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