a
Given \({P}_{{A}}^{\circ}=90\, {~mm}\, {Hg}\), at \(25^{\circ} {C}\)

\({P}_{{B}}^{\circ}=15\, {~mm} \,{Hg}\)

\(\text { and }{X}_{{A}}=0.6,{X}_{{B}}=0.4\)

\({P}_{{T}}={X}_{{A}} {T}_{{A}}^{\circ}+{X}_{{B}} {P}_{{B}}^{\circ}\)

\(=(0.6 \times 90)+(0.4 \times 15)\)

\(=54+6=60\, {~mm}\)

Now mol fraction of \(B\) in the vapour phase

i.e. \(Y_{B}=\frac{P_{B}}{P_{T}}=\frac{X_{B} P_{B}^{\circ}}{60}=0.1=1 \times 10^{-1}\)

Therefore: \(x=1\)