A) 0.70
B) 0.50
C) 0.60
D) 0.80
E) 0.40
Correct Answer: A
Solution :
According to Raoults law. Relative lowering of vapour pressure\[\propto \]mole fraction of solute Thus, mole fraction of solute = 0.0125 Mole fraction of a solute is related to the molality by the following expression. \[\left( \frac{1}{X}-1 \right)=\frac{1000}{{{m}_{B}}\times m}\] where,\[X=\]mole fraction of solute \[{{m}_{B}}=\]molecular weight of solvent \[m=\]molality \[\left( \frac{1}{0.0125}-1 \right)=\frac{1000}{18\times m}\] \[m=\frac{12.5}{(1-0.0125)\times 18}\] \[=\frac{12.5}{17.775}\] \[=0.70\]
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