(i) 0.24 g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature. (i) What is a vapour pressure of liquid?

(i) 0.24 g of a gas dissolves in 1 L of water at 1.5 atm pressure. Cal

1.	(i) 0.24 g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature. (i) What is a vapour pressure of liquid?
Answer» Solution :`P_(sides)=K_(H)x_("acid in solutions")` At PRESSURE 1.5 atm, `P_(1)=K_(H)x_(1)`.....(1) At pressure 6.0 atm, `P_(2)=K_(H)x_(2)`.....( 2) Dividing EQUATION (1) by (2) We get`(p_(1))/(p_(2))=(x_(1))/(x_(2))RARR(1.5)/(6.0)=(0.24)/(x_(2)0` Therefore`x_(2)=(0.24xx6.0)/(1.5)=0.96g//L` (ii) 1. The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature. 2. The relative lowering of vapour pressure is defined as the RATIO of lowering of vapour pressure to vapour pressure of PURE solvent. Relative lowering of vapour pressure = `(P_(solvent)-P_(solution))/(P_(solvent))`