The vapour pressure of a dilute aqueous solution of glucose is 740 mm of mercury at 373 K . The mole fraction of the solute is

Question

The vapour pressure of a dilute aqueous solution of glucose is 740 mm of mercury at 373 K . The mole fraction of the solute is (A)  (1/20)  (B)  (1/38)  (C)  (1/76)  (D)  (1/740)

Tardigrade Admin · Accepted Answer

At STP (273 K ) the vapour pressure of aqueous solution of glucose =760 mm Hg At 373 K, the V P=740 mm Hg We know that, Relative lowering of V P propto mole fraction of solute (po-p/po)=XA (760-740/760) =XA (20/760)=XA ∴ XA=(2/76)=(1/38)

Q. The vapour pressure of a dilute aqueous solution of glucose is $740 mm$ of mercury at $373 K$ . The mole fraction of the solute is

$\frac{1}{20}$

$\frac{1}{38}$

$\frac{1}{76}$

$\frac{1}{740}$

Q. The vapour pressure of a dilute aqueous solution of glucose is 740mm of mercury at 373K . The mole fraction of the solute is

201​

381​

761​

7401​

At STP (273K) the vapour pressure of aqueous solution of glucose =760mmHg At 373K, the VP=740mmHg We know that, Relative lowering of VP∝ mole fraction of solute popo−p​=XA​760760−740​ =XA​76020​=XA​ ∴XA​=762​=381​

Q. The vapour pressure of a dilute aqueous solution of glucose is $740 mm$ of mercury at $373 K$ . The mole fraction of the solute is

$\frac{1}{20}$

$\frac{1}{38}$

$\frac{1}{76}$

$\frac{1}{740}$