750 mL of nitrogen are collected over water at `25^@C` and 740 mm pressure. If the aqueous tension at this temperature is 23.8 mm Hg, calculate the mass of the dry gas.

Explanation

According to Dalton.s law of partial pressure, `P_("gas")= P_("obs.)`- aqueous tension = `740-23.8 = 716.2mm Hg`. The given gas thus occupies a volume of 750 mL at `25^(@)` and 716.2 mm Hg.
According to the ideal gas equation, `PV= nRT= (m)/(M) RT`
`m= (PVM)/(RT)`
Here, P= 716.2 mm Hg `= (716.2)/(760)= 0.9424` atm.
V= 750 mL = 0.750L
T= `25 + 273 = 298K`
`M= 28 and R= 0.0821 L " atm " K^(-1) mol^(-1)`
`m= (0.9424 xx 0.750 xx 28)/(0.0821 xx 298)= 0.809g`
Hence the mass of the given gas= 0.809g