Given :
A mixture of water and acetone at 756 mm boils at 70.0°C.
The vapor pressure of acetone is 1.54 atm at 70.0°C, while the vapor pressure of water is 0.312 atm at the same temperature.
To Find :
The percentage composition of the mixture.
Solution :
By Raoult's law :
[tex]P=x_{acetone}P^o_{acetone}+x_{water}P^o_{water}\\\\x_{acetone}1.58+x_{water}0.312=\dfrac{756}{760}=0.995\ atm\\\\1.58x_a+0.312x_w=0.995[/tex]......( 1 )
Also , [tex]x_a+x_b=1[/tex] ......( 2 )
Solving equation 1 and 2 , we get :
[tex]x_a=0.54\ and \ x_w=0.46[/tex] .
Mass of acetone ,
[tex]m_a=x_a\times MM_a\\\\m_a=0.54\times 58\\\\m_a=31.32\ g[/tex]
Mass of water ,
[tex]m_w=x_w\times MM_w\\\\m_w=0.46\times 18\\\\m_a=8.28\ g[/tex]
[tex]\%water =\dfrac{8.28}{8.28+31.32}\times 100\\\\\%water =20.9\%\\\\\%acetone =79.1\%[/tex]
Hence , this is the required solution.
