Question

4. The vapour pressure of an aqueous solution of cane sugar (molecular mass 342) is 756mm at 373K. How many grams of sugar are present in 1 kg of water?​

Answer 1

we know that

• vapour pressure of pure solvent(water) =760

given

• The molecular mass of cane sugar is 342
• vapour pressure of the solution is 756mm at 373k.

solution

• Xsolute$Xsolute= \frac{760-756}{760} =\frac{4}{760}$ (as x solute $\frac{moles of solute}{total moles of solvent}$
• $\frac{n solute}{ns solvent} =\frac{4}{756}$
• $\frac{weight of solute}{342}/\frac{1000}{18}= \frac{4}{756}$

   weight of solute=$\frac{4*342*1000}{756*18} = 100.5gm$