Question

A mixture of $H_{2}O$ ,$C{O}_2$ and $N_2$ are trapped in a glass apparatus with a volume of 0.731 mL. The pressure of total mixture was 1.74mm of Hg at ${23}^o$. The sample was transferred to a bulb in contact with dry ice(${75}^o$) so that $H_{2}O\left(v\right)$ are frozen out. When the sample returned to the normal value of temperature, the pressure was 1.32mm of Hg. The sample was then transferred to bulb in contact with liquid $N_2(-{75}^o)$ to freeze out $C{O}_2$. In the measured volume, the pressure was 0.53mm of Hg at original temperature. Mole of $N_2$ in mixture are $y\ast {10}^{-8}$.
So, the value of y is......(as nearest integer)

Answer 1

Volume of container =0.731mL

Temperature =23+273=296K

Initial pressure $=P_{H_{2}O}+P_{C{O}_2}+P_{N_2}$=1.74mm.............(i)

At ${-75}^o$ $H_{2}O$ is frozen out and the mixture on returning at ${23}^o$ has pressure due to $C{O}_2$ and $N_2$ .

$P=P_{C{O}_2}+P_{N_2}=1.32$ mm.........................(ii)

At${-95}^o$ $C{O}_2$ is also frozen out and the mixture on returning at ${23}^o$ has pressure due to $N_2$ alone

$P_{N_2}=$0.53mm.........................(iii)

By Eqs (ii) and (iii),$P_{C{O}_2}=1.32-0.53=0.79$mm

By Eqs (i) and (ii),$P_{H_{2}O}$=1.74-1.32=0.42mm

Now using PV=nRT for each gas separately

For $N_2$; n=$\frac{PV}{RT}$=$\frac{0.53\ast 0.731}{760\ast 0.0821\ast 296\ast 1000}$=$2.1\ast {10}^{-8}$

For $C{O}_2$; n=$\frac{PV}{RT}$=$\frac{0.79\ast 0.731}{760\ast 0.0821\ast 296\ast 1000}$ = $3.1\ast {10}^{-8}$

For $H_{2}O$; n=$\frac{PV}{RT}$=$\frac{0.42\ast 0.731}{760\ast 0.0821\ast 296\ast 1000}$= $1.7\ast {10}^{-8}$

So, value of y is 2.(as nearest integer)