c
Let vapour pressure of \(A=P_A^o\)

vapour pressure of \(B=P_B^o\)

In first solution,

Mole fraction of \(A\,({x_A})\, = \,\frac{1}{{1 + 2}}\, = \,\frac{1}{3}\)

Mole fraction of \(B\,({x_B})\, = \,\frac{2}{{1 + 2}}\, = \,\frac{2}{3}\)

According to Raoult's law, 

Total vapour pressure

\( = \,250\, = \,P_A^o{x_A}\, + \,P_B^o{x_B}\)

\(\,250\, = \,\frac{1}{3}P_A^o + \,\frac{2}{3}P_B^o\)   ....... \((i)\)

In second solution

Mole fraction of \(A\,({x_A})\, = \,\frac{2}{{2\, + \,2}}\, = \,\frac{2}{4}\, = \frac{1}{2}\)

Mole fraction of \(B({x_B})\,\, = \,\frac{2}{4}\, = \frac{1}{2}\)

\(\therefore \) Total vapour pressure

\( = \,300\, = \,P_A^o{x_A}\, + \,P_B^o{x_B}\)

\(300\, = \,\frac{1}{2}P_A^o\, + \,\frac{1}{2}P_B^o\)   ..... \((ii)\)

Multiplying equation \((i)\) by \(\frac {1}{2}\) and equation \((ii)\) by \(\frac {1}{3}\)

\(\frac{1}{6}P_A^o\, + \,\frac{2}{6}P_B^o\, = \,125\)

\(\frac{1}{6}P_A^o\, + \,\frac{1}{6}P_B^o\, = \,100\)

\(\frac{1}{6}P_B^o\, = \,25\)

\(P_B^o\, = \,25\, \times \,6\, = \,150\,mm\,Hg\)

On substituting value of \(P_B^o\) in equation \((ii)\) we get

\(300\, = \,P_A^o\, \times \,\frac{1}{2}\, + \,150\, \times \frac{1}{2}\)

\(\,P_A^o\, = \,450\,mm\,Hg\)