Question

The vapour pressure of a pure liquid at 298K is 4x10^4 N/m2. When a non-volatile solute is dissolved, the vapour pressure becomes 3.65x10^4 N/m2 Calculate relative lowering vapour pressure and mole- fraction of solute dissolved in that liquid.​

Answer 1

Given:

vapour pressure of a pure liquid =$p^{0}=4*10^{4}N/m^{2}$

Vapour pressure of solution=$p=3.65*10^{4}N/m^{2}$

Temperature T=298K

To find: relative pressure=$p/p^{0} =?$

lowering of  vapour pressure$=p^{0} -p=?$

Relative lowering of pressure=$(p^{0}-p)/p^{0}=?$

relative  vapour pressure=p/p0=$(3.65*10^{4} )/4*10^{4}$

                                                    =0.9125

lowering of vapour pressure=$4*10^{4}-3.65*10^{4}\\=0.35*10^{3} \\=3.5*10^{4}$

relative lowering of vapour pressure=(p0-p)/p0=3.5×10³/4*10^4

                                                             =0.0875