18 g of glucose is dissolved in 178.2 g of water. The vapour pressure of the solution at 100°C is (vapour pressure of pure water at 100°C is 760 mm Hg)

Question

18 g of glucose is dissolved in 178.2 g of water. The vapour pressure of the solution at 100°C is (vapour pressure of pure water at 100°C is 760 mm Hg) (A) 767.6 mm Hg (B) 760 mm Hg (C) 752.4 mm Hg (D) 725.4 mm Hg

Tardigrade Admin · Accepted Answer

Vapour pressure (p°) of pure water at 100° C is =760 mm Hg From relative lowering of vapour pressure (p°-ps/p°) =(W2/m2) × (m1/W1) (760-ps/760) =(18/180) × (18/178.2)=(1/99) 760-ps =(760/99) or 760-ps =7.6 ⇒ ps =752.4 mm Hg

Q. $18 g$ of glucose is dissolved in $178.2 g$ of water. The vapour pressure of the solution at $10 0^{\circ} C$ is (vapour pressure of pure water at $10 0^{\circ} C$ is $760 mm H g$ )

$767.6 mm H g$

$760 mm H g$

$752.4 mm H g$

$725.4 mm H g$

$745.2 mm H g$

Q. 18g of glucose is dissolved in 178.2g of water. The vapour pressure of the solution at 100∘C is (vapour pressure of pure water at 100∘C is 760mmHg)

767.6mmHg

760mmHg

752.4mmHg

725.4mmHg

745.2mmHg