Answer:

The relative lowering of vapour pressure is given by the following expression. 
$\frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -{\mathrm{p}}_{\mathrm{solution}}}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{n_2}{n_1+n_2}$

Where ${\mathrm{p}}_{\mathrm{solvent}}^0$ is the vapour pressure of pure solvent, P_solution  is the vapour pressure of solution containing dissolved solute, n₁ is the number of moles of solvent and n₂ is the number of moles of solute.

For dilute solution n₂<<n₁, therefore the above expression reduces to 

$$\begin{gathered} \frac{{\mathrm{p}}_{\mathrm{solvent}}^0- {\mathrm{p}}_{\mathrm{solution}}}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{n_2}{n_1} \\ = \frac{w_2 x{M}_1}{M_2 x w_1} .....\left(A\right) \end{gathered}$$

Where w₁ and w₂ are the masses and M₁ and M₂ are the molar masses of solvent and solute respectively.

We are given that

W₁ =30 g

W₂ =30g

p_solution =2.8 kpa

p⁰_solvent =? And M₂=?

$$\begin{gathered} \mathrm{subtituting} \mathrm{these} \mathrm{values} \mathrm{in} \mathrm{relation} \\ \mathrm{we} \mathrm{get}, \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.8}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{30 \mathrm{x}18}{{\mathrm{M}}_2 \mathrm{x}90} \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.8}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{6}{{\mathrm{M}}_2 } \left(1\right) \\ \mathrm{similarly} \mathrm{for} \mathrm{second} \mathrm{case} \mathrm{we} \mathrm{have} \mathrm{the} \mathrm{following} \mathrm{value}. \\ {\mathrm{w}}_2 =90 \mathrm{g} \\ {\mathrm{w}}_1 =90+18 =108\mathrm{g} \\ {\mathrm{p}}_{\mathrm{solution}} =2.9 \mathrm{kpa} \\ \mathrm{therefore} \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.9}{{\mathrm{p}}_{\mathrm{solvent}}^0} =\frac{30 \mathrm{x}18}{{\mathrm{M}}_2 \mathrm{x}108} \left(2\right) \\ \mathrm{dividing} 1 \mathrm{and} 2 \mathrm{we} \mathrm{get} \\ \frac{{\mathrm{p}}_{\mathrm{solvent}}^0 -2.8}{{\mathrm{p}}_{\mathrm{solvent}}^0-2.9} =\frac{6}{5} \\ \mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{water} \mathrm{at} 298 \mathrm{k} \mathrm{is} 3.4\mathrm{kpa} \\ \mathrm{substituting} \mathrm{the} \mathrm{vaue} \mathrm{of} {\mathrm{p}}_{\mathrm{solvent}}^0 \mathrm{in} 1 \mathrm{we} \mathrm{get} \\ \frac{3.4 -2.8}{3.4} =\frac{6}{{\mathrm{M}}_2} \\ = 34 \mathrm{gram} \\ \mathrm{Therefore} \mathrm{mass} \mathrm{of} \mathrm{solute} \mathrm{is} 34 \mathrm{g} . \end{gathered}$$

Answer:

AS both the solution are  isotonic they should have same concentrationin mole/litre
for sucrose solution concentration =4g/100cm³
                                                   = 40g/Litre

molar mass of sucrose C ₁₂H₂₂O₁₁ = 342
        therefore we get =40/342 moles/litre

For unknown substance Let N be the molecular mass then concentration = 3g/100cm³ 
                                    = 30g/Litre

                                   = 30/N moles/Litre
comparing the both equation
30/N = 40/342

N=(30 x 342)/40

N= 256.5
Molecular mass of oragnic compound

Answer:

$\frac{\mathrm{p}°-\mathrm{p}}{\mathrm{p}°}$ = x_urea

Let the mass of solution be 100g 
 therefore mass of urea = 10 g
 molecular mass of urea = 60 g
 
x_urea = 10/60 =1/6

molecular mass of water = 18g
x_water = 90/18 =5

$$\begin{gathered} {\mathrm{x}}_{\mathrm{urea}} =\frac{ \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{urea} }{\mathrm{total} \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{solution}} \\ {\mathrm{x}}_{\mathrm{urea}} =\frac{1/6}{1/6+5} = \frac{1/6}{31/6} \\ {\mathrm{x}}_{\mathrm{urea}} = 1/31 \\ \mathrm{using} \mathrm{above} \mathrm{formula} \\ \frac{ \\ 55.3-\mathrm{p}}{55.3} = \frac{1}{31} \\ 55.3-\mathrm{p} = \frac{55.3}{31} \\ \mathrm{p}= 55.3-\frac{55.3}{31} \\ 55.3\left(1-\frac{1}{31}\right) =\mathrm{p} \\ \mathrm{p} =55.3 \times \frac{30}{31} \\ \mathrm{p}= 53.52\mathrm{mm} \mathrm{Hg} \end{gathered}$$

W_B =5.67g

W_A =25.2g

M_B=180g/mol

M_A=18g/mol

N_B= 5.67/180 =0.0315 mol glucose

N_A=25.2/18 =1.40 mol water

$$\begin{gathered} \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{glucose}, \\ {\mathrm{x}}_{\mathrm{B}} =\frac{0.0315}{1.40 +0.0315} =0.022 \\ \mathrm{we} \mathrm{know} \mathrm{that} \\ \mathrm{lowering} \mathrm{of} \mathrm{V}.\mathrm{P} =\frac{ {\mathrm{p}}^0 -\mathrm{p} }{{\mathrm{p}}^0} \\ {\mathrm{x}}_{\mathrm{B}} =23.8 \mathrm{x}0.022 =0524 \mathrm{mmHg} \\ \frac{{\mathrm{p}}^0-\mathrm{p}}{{\mathrm{p}}^0} ={\mathrm{x}}_{\mathrm{B}} \\ \mathrm{p}={\mathrm{p}}_{\mathrm{A}}^0 -({\mathrm{p}}^0-\mathrm{p}) \\ =23.8 -0.524 \\ =23.3\mathrm{mm} \mathrm{Hg} \end{gathered}$$