$$\eqalign{ & {\text{No}}{\text{. of moles of}}\,\,NaOH = \frac{{10}}{{40}} = 0.25\,mol \cr & M = \frac{{0.25 \times 1000}}{{500}} = 0.5\,\,mol\,\,{L^{ - 1}} \cr} $$

$$\eqalign{ & {M_B} = \frac{{{K_b} \times {W_B}}}{{\Delta {T_b} \times {W_A}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{0.52 \times 12.5}}{{0.80 \times 0.185}} \cr & \,\,\,\,\,\,\,\, = 43.92\,g\,mo{l^{ - 1}}\,\,\left( {\because \,\,185\,g = 0.185\,kg} \right) \cr} $$

Let the mass of methane and oxygen $${\text{ = }}m{\text{ }}gm.$$
Mole fraction of $${O_2}$$
$$\eqalign{ & = \frac{{{\text{Moles of}}\,{O_2}}}{{{\text{Moles of}}\,{O_2} + {\text{Moles of}}\,C{H_4}}} \cr & = \frac{{\frac{m}{{32}}}}{{\frac{m}{{32}} + \frac{m}{{16}}}} \cr & = \frac{{\frac{m}{{32}}}}{{\frac{{3m}}{{32}}}} \cr & = \frac{1}{3} \cr} $$
Partial pressure of $${O_2} = $$  Total pressure $$ \times $$  mole fraction of $${O_2},\,{P_{{O_2}}} = P \times \frac{1}{3} = \frac{1}{3}P$$

$$\eqalign{ & \frac{{{p^ \circ } - {p_A}}}{{{p^ \circ }}} = \frac{{{n_B}}}{{{n_A}}} \cr & \frac{{2 - 1}}{2} = \frac{{{W_B}}}{{{M_B}}} \times \frac{{{M_A}}}{{{W_A}}} \cr & \Rightarrow \frac{1}{2} = \frac{1}{{{M_B}}} \times \frac{{200}}{{20}} \cr & \Rightarrow {M_B} = \frac{{200 \times 2}}{{20}} = 20 \cr} $$

The Tyndall effect is observed only when following conditions are satisfied :
The diameter of the dispersed particles is much smaller than the wavelength of the light used.
The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

$$\eqalign{ & {\text{Moles of urea}} = \frac{{12}}{{60}} = 0.2 \cr & {\text{Moles of sucrose}} = \frac{{68.4}}{{342}} = 0.2 \cr} $$
Both are non electrolyte hence lowering of $$V.P.$$  will be same.

$$\eqalign{ & {\text{For an ideal solution,}} \cr & \Delta {H_{sol.}} = \Delta {H_1} + \Delta {H_2} + \Delta {H_3} \cr} $$

Since, component having higher vapour pressure will have higher percentage in vapour phase. Benzene has vapour pressure $$12.8$$ $$kPa$$  which is greater than toluene $$3.85$$ $$kPa.$$
Therefore, the vapour will contain a higher percentage of benzene.

According to Raoult’s law, the relative lowering of vapour pressure is equal to the mole fraction of solute, i.e.
$$\eqalign{ & \frac{{{p^ \circ } - p}}{{{p^ \circ }}} = {\chi _B} \cr & {\chi _B} = {\text{mole fraction of solute}} \cr} $$

No. of moles of glucose $$ = \frac{{10}}{{180}} = 0.0555\,mol$$
No. of moles of water $$ = \frac{{90}}{{18}} = 5\,mol$$
Number of moles of solution $$ = 5.0555\,mol$$
Mole fraction of glucose
$$\eqalign{ & \, = \frac{{{\text{No}}{\text{.}}\,\,{\text{of}}\,\,{\text{moles}}\,\,{\text{of}}\,\,{\text{glucose}}}}{{{\text{No}}{\text{.}}\,\,{\text{of}}\,\,{\text{moles}}\,\,{\text{of}}\,\,{\text{solution}}}} \cr & = \frac{{0.0555}}{{5.0555}} \cr & = 0.01 \cr} $$