The volume of a given mass of a gas is 919 mL in dry state at STP. The same mass when collected over water at `15^(@)C` and 750 mm pressure occupies a volume of 1 L. Find out the vapor pressure of water at `15^(@)C`.

Explanation

Givne : `V_(1)= 919 mL, P_(1)= 760 mm, T_(1)= 273K, V_(2)= 1000mL, T_(2)= 273+15= 288 K`
Pressure of the dry gas, `P_(2)= (P_(1)V_(1)T_(2))/(V_(2)T_(1))= (760 mm xx 919 mL xx 288 K)/(1000 mL xx 273 K)= 736.8 mm`
Vapour pressure of water= Pressure of the moist gas-Pressure of the dry gas `= 750-736.8= 13.2 mm`.