The vapour pressure of two liquid P and Q are 80 torr and 60 torr respectively. The total vapour pressure obtained by mixing `3` moles of P and 2 mole of Q would be

Options

A. 72 torr
B. 20 torr
C. 68 torr
D. 140 torr

Answer

1

Explanation

We are given
According to Raoult's law `P_("total")=P_(P)^(0)chi_(P)+P_(Q)^(0)chi_(Q)`
`P_(P)^(0)`=80 torr
`P_(Q)^(0)`=60 torr
`n_(p)=3 mol`
`n_(Q)=2 mol`
Thus
`chi_(P)=n_(P)/(n_(P)+n_(Q))=(3 mol)/(3 mol+ 2 mol)`
`=0.6`
`chi_(Q)=1-chi_(P)=1-0.6=0.4`
Using these results, we have
`P_("total")=(80 "torr")(0.6)+(60 "torr")(0.4)`
=48 torr+ 24 torr
=72 torr