A) \[\frac{3}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{4}{3}\]
Correct Answer: A
Solution :
The relationship between molar masses of two solvents is \[{{M}_{X}}=\frac{3}{4}{{M}_{Y}}.........(1)\] The relative lowering of vapour pressure of two solutions is But, the relative lowering of vapour pressure of solution is directly proportional to the mole fraction of solute. \[{{M}_{x}}\times \frac{5}{1000}=m\times {{M}_{Y}}\times \frac{5}{1000}.......(2)\] Substitute equation (1) in equation (2). \[\frac{3}{4}\times {{M}_{Y}}\times \frac{5}{1000}=m\times {{M}_{Y}}\times \frac{5}{1000}\] \[m=\frac{3}{4}\] Note: 5 molal solution means 5 moles of solute are dissolved in 1 kg (or 1000 g) of solvent. The number of moles of solvent \[=\frac{1000g}{M}\] \[=\frac{5}{1000/M}\] \[=M\times \frac{5}{1000}\]
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