Answer
In the container liquid ethanol present $=0.44\text{ g}$
Work Step by Step
When both the liquid and vapor phases exist in equilibrium ,the pressure exerted by the pure compound is called vapor pressure .
The vapor of ethanol
$=349mmHg$
$=349mmHg\times1atm/760mmHg$
$=0.4592$
Temperature $T=60^{\circ}C=(273+60)K=333K$
Volume $V=400mL=0.400L$
Assuming ethanol vapor as an ideal gas, $PV=nRT$
$n=PV/RT=(0.4592atm\times0.400L)/0.08206LatmK^-$$^1mol^-$$^1333K$
$=6.722\times10^-$$^1mol$
Mole of ethanol present $=0.75g/46.07g$ $mole^-$$^1=0.01628moles$
So, in the container liquid ethanol present
$=(0.06128-0.00722)mole\times46.07g/mole$
$=0.44g$
