Question

An aqueous solution of glucose is made by dissolving 10g glucose in 90g of water at 303K. If vapour pressure of pure water at 303K is 32.8 mm of Hg, what would be vapour pressure of solution?

Answer 1

The solution is in the attachment.

Hope it helps.......

Answer 2

Answer : The vapor pressure of solution is, 23.324 torr

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent (water)  = 32.8 mmHg

$p_s$ = vapor pressure of solution = ?

$w_2$ = mass of solute  (glucose) = 10 g

$w_1$ = mass of solvent  (water) = 90 g

$M_1$ = molar mass of solvent (water) = 18 g/mole

$M_2$ = molar mass of solute (glucose) = 180 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{32.8-p_s}{32.8}=\frac{10\times 18}{180\times 90}$

$p_s=32.44mmHg$

Therefore, the vapor pressure of solution is, 32.44 mmHg