Vapour pressure of water decreases from one atmosphere to one bar by dissolving glucose. 
(1 atm = 1.01 bar). Which statement(s) is(are) correct ? (P_bar = 1atm).

$\begin{array}{l}\frac{{\mathrm{P}}^0-\mathrm{P}}{{\mathrm{P}}^0}={\mathrm{X}}_{\mathrm{B}};{\mathrm{X}}_{\mathrm{B}}=\frac{(1.0-1)}{1.01}=\left(\frac{0.01}{1.01}\right)=0.0099\\ \frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{B}}}=\left(\frac{1}{0.0099}-1\right)\Rightarrow \frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{B}}}=\left(\frac{1-0.0099}{0.0099}\right)=100\\ \Rightarrow \frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}=\frac{1}{100}=0.01\Rightarrow 1\mathrm{\%}\text{ solute by moles }\\ \Rightarrow \frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}=\frac{1\text{ mole }}{100\text{ moles }}\Rightarrow {\mathrm{n}}_{\mathrm{B}}=1\text{ mole }\\ \Rightarrow {\mathrm{W}}_{\mathrm{B}}=180\text{ gm }\Rightarrow {\mathrm{n}}_{\mathrm{A}}=100\text{ mole }\Rightarrow {\mathrm{W}}_{\mathrm{A}}=1800\mathrm{gm}\\ \Rightarrow {\mathrm{wt}}^2\text{ of glucose }=\frac{180}{1800+180}\times 100\\ =\frac{180\times 100}{1980}=9.09\mathrm{\%}\end{array}$