Question

The vapour pressure of an aqueous solution of glucose is 750 mm Hg at 373 K. The molarity and mole fraction of solute are :

• A. $0.731M,0.013$
• B. $0.441M,0.013$
• C. $0.741M,0.014$
• D. None of the above

Answer 1

The correct option is A $0.731M,0.013$

The expression for the relative lowering in the vapor pressure and the mole fraction of solute is $\frac{\Delta P}{P}=\frac{n}{N}$
For 1 L of solution $\frac{10}{760}=\frac{n}{55.56}$

Thus, the number of moles of solute $n=0.731$

Thus, the mole fraction of solute $=\frac{0.731}{55.56}=0.013$

The molarity of the solution is $\frac{0.731}{1}=0.731M$