Question

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.
Calculate the percentage composition of the mixture

Answer 1

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 :

`P=x_(acetone)P^o_(acetone)+x_(water)P^o_(water)\\\\x_(acetone)1.58+x_(water)0.312=(756)/(760)=0.995\ atm\\\\1.58x_a+0.312x_w=0.995`......( 1 )

Also ,
`x_a+x_b=1` ......( 2 )

Solving equation 1 and 2 , we get :

`x_a=0.54\ and \ x_w=0.46` .

Mass of acetone ,

`m_a=x_a* MM_a\\\\m_a=0.54* 58\\\\m_a=31.32\ g`

Mass of water ,

`m_w=x_w* MM_w\\\\m_w=0.46* 18\\\\m_a=8.28\ g`

`\%water =(8.28)/(8.28+31.32)* 100\\\\\%water =20.9\%\\\\\%acetone =79.1\%`

Hence , this is the required solution.