A) 0.70
B) 0.50
C) 0.90
D) 0.80
E) 0.60
Correct Answer: A
Solution :
Relative lowering of vapour pressure = mole fraction of solute (Raoults law) \[\frac{P-{{P}_{S}}}{P}={{X}_{2}}\] \[\frac{P-{{P}_{S}}}{P}=\frac{wM}{mW}\] where w = wt. of solute M = mol. wt. of solvent m = mol. wt of solute W = wt. of solvent \[0.0125=\frac{wM}{mW}\] Or, \[\frac{w}{mW}=\frac{0.0125}{18}=0.00070\] Hence, molality \[=\frac{w}{mW}\times 1000=0.0007\times 1000=0.70\]
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