Molarity of solution = \(\frac { 0.025\ mole }{ 0.200\ L } =0.125\ M\)

\(d=\frac { 118001.5 }{ 94500 } =1.2487 \ g{ L }^{ -1 }\)

\({ P }_{ 1 }=Antilog\ \bar { 1 } .1928=0.1559\ atm\)

72 g

\(\pi =CRT\)
\(\therefore\) In the first case, \(4.98=\frac { 36 }{ 180 } \times R\times 300=60R\)
In the second case,\(1.52=C\times R\times 300\)
Dividing (ii) by (i), we get C = 0.061 M.

At any fixed temperature, the vapour phase is always richer in the more volatile component as compared to the solution phase. In other words, mole fraction of the more volatile component is always greater in the vapour phase than in the solution phase. Alternatively, vapour phase is relatively richer in the component whose addition to the liquid mixture results in an increase in the total vapour pressure.

Henry's law is sometimes expressed in another way as follows:
The volume of the gas dissolved, measured at the pressure used, is independent of pressure.
This can be easily shown as under :
Suppose volume of the liquid taken = V cc
Pressure of the gas = p mm
Mass of the gas dissolved = m gm
Suppose, volume of m gm of the gas at pressure p mm = v cc
Now, suppose the pressure is doubled, i.e., it becomes = 2 p
∴ Mass of the gas dissolved will be = 2 m grams
But volume of m grams of the gas at pressure p = v cc
∴ Volume of 2 m grams of the gas at pressure p = 2 v cc
or Volume of 2 m grams of the gas at pressure 2 p = v cc
(by Boyle's law because temperature is kept constant)
Thus, the volume of the gas dissolved at pressure 2 p is again the same, viz., v cc.