Question

# A solution containing  6.35 g of a nonelectrolyte dissolved in 500 g of water freezes at $$-0.465^{\circ}C$$ Determine the molecular weight of the solute. $$[K_{f} for \ water 1.86^{\circ}C/m]$$

A
25.4 g/mol
B
50.8 g/mol
C
76.2 g/mol
D
90.2 g/mol

Solution

## The correct option is B $$50.8$$ g/molThe depression in the freezing point $$=\Delta K_f = 0 - (-0.465) = 0.465$$$$\Delta T_f = K_f \times m$$$$0.465 = 1.86\ m$$$$m = 0.25\ m$$Let $$M$$ be the molecular weight of the solute.Number of moles is the ratio of mass to molecular weight.Number of moles of solute $$=\dfrac {6.35}{M}$$The molality of a solution is the number of moles of solute in $$1000$$ g of water.Molality $$=\dfrac {6.35 \times 1000}{M \times 500} = 0.25$$$$\implies M = 50.8 \: g/mol$$Chemistry

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