Subjective Type

The freezing point of a solution containing 5.85 g of $$NaCl$$ in 100g of water is $$-3.348^o$$C. Calculate van't Hoff factor 'i' for this solution. What will be the experimental molecular weight of NaCI? ($$K_f$$ for water = $$1.86 K kg mol^{-1}$$, at. wt. Na = 23, Cl = 35.5)

Solution

$$w_{ NaCl }$$=5.85 g; $$w_{ H_2O }$$= 100 g; $$\Delta T_f$$= 3.348 K; $$K_f$$=1.86 Kkg/mol
As, $$\Delta { T }_{ f }=i{ K }_{ f }m\\ m=\frac { { w }_{ NaCl } }{ { M }_{ NaCl } } \times \frac { 1000 }{ { w }_{ { H }_{ 2 }O } } \\ m=\frac { 5.85 }{ 58.5 } \times \frac { 1000 }{ 100 } =1\\Therefore,\ i=\frac { 3.348 }{ 1.86 } =1.8\\ Also,\ i=\frac { { Normal\ molecular\ wt } }{ { Abnormal\ molecular\ wt } } \\ 1.8=\frac { 58.5 }{ { M } } \\ \therefore M=32.5\ g/mol$$
Vant Hoff factor is $$1.8$$ and experimental molecular weight of $$NaCl$$ is 32.5 g/mol.

SIMILAR QUESTIONS

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