Question

the vapour pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 k. the mole fraction of solute is:

Answer 1

Given in the question :-

Temperature = 373 Kvapour pressure H2O , p°= 760 mm
$p^s$ = 750 mm

From the Rault law 

$(p^0-p^s)/p^0$ 

Now mole fraction of the solute
Put the values in the formula 
$(p^0-p^s)/p^0$ 

$\frac{760-750}{760}$
$10/760$
0.013 

Hence the mole fraction is 0.013


Hope it Helps :-) 

Answer 2

Answer:The mole fraction of solute is 0.013.

Explanation:

Given:

Vapour pressure of a dilute aqueous solution of glucose at 373 K( at $100^o C$) =$p_s$=750 mm Hg

Vapour pressure of pure solvent that is water(at $100^o C$)=$p^o$= 760 mm Hg

According to Raoult's law for non volatile solutes in solution:

$\frac{p^o-p_s}{p^o}=X_{solute}$

On substituting the given values:

$X_{glucose}=\frac{p^o-p_s}{p^o}=\frac{760 -750}{760}=0.013$

The mole fraction of solute is 0.013.