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.

750 mL of nitrogen are collected over water at 25^@C and 740 mm pressu

1.	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.
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :According to Dalton.s law of partial pressure, <br/> `P_(gas) = P_(obs). - " aqueous <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> " = 740 - 23.8 = 716.2 mm Hg` <br/>The given gas thus occupies a volume of <a href="https://interviewquestions.tuteehub.com/tag/750-335306" style="font-weight:bold;" target="_blank" title="Click to know more about 750">750</a> mL at `25^@C " and " 716.2 mm Hg`. <br/> According to the ideal gas equation, <br/> `PV =nRT= m/M RT` <br/> `m=(PVM)/(RT)` <br/> In the present case, <br/> `P =716.2 mm = (716.2)/<a href="https://interviewquestions.tuteehub.com/tag/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a> =0.9424` atm <br/> `V=750 mL = 0.750 L, "" T=25+273 =298 K` <br/> `M = 28 " and " R =0.0821 L " atm " K^(-1) mol^(-1)` <br/> `m=(0.9424xx0.750xx28)/(0.0821xx298) = 0.809 g` <br/> Hence, the mass of the given gas = 0.809 g</body></html>