$$\begin{gathered} \mathrm{At} \mathrm{the} \mathrm{boiling} \mathrm{point} \mathrm{of} \mathrm{pure} \mathrm{solvent} \mathrm{the} \mathrm{vapour} \mathrm{preesure} \mathrm{is} 1 \mathrm{atm} \\ {\mathrm{p}}^0 =1 \mathrm{Atm} \\ \mathrm{p} =0.096 \\ \frac{{\mathrm{p}}^0-\mathrm{p}}{{\mathrm{p}}^0} =0.004 \\ 2\% \mathrm{solution} \mathrm{means} 2\mathrm{g} \mathrm{solute} \mathrm{in} 100\mathrm{g} \mathrm{solution},\mathrm{so} \\ {\mathrm{w}}_{\mathrm{B}} =2\mathrm{g} \\ {\mathrm{w}}_{\mathrm{A}} =98\mathrm{g} \\ {\mathrm{M}}_{\mathrm{A}} =18\mathrm{g}/\mathrm{mol} \\ \mathrm{we} \mathrm{know} \mathrm{that} \\ {\mathrm{M}}_{\mathrm{B}} =\frac{{\mathrm{w}}_{\mathrm{B}}{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{w}}_{\mathrm{A}}{\displaystyle \frac{\left({\mathrm{p}}^0-\mathrm{p}\right)}{{\mathrm{p}}^0}}} \\ =\frac{2 \mathrm{x}18}{98 \mathrm{x} 0.004 } =45.5\mathrm{g}/\mathrm{mol} \end{gathered}$$

Answer:

The molality of the cane sugar, m= 0.1539m
depression in freezing point $∆{\mathrm{T}}_{\mathrm{f}}$ = 273.15-271
                                                 =2.15K

Since $∆{\mathrm{T}}_{\mathrm{f}}$ = K_fm
 or K_f = $∆{\mathrm{T}}_{\mathrm{f}}$/m  = 2.15k/0.1539m  = 13.97K/m

Now the weight of glucose , W₂ = 5g
molecular mass of glucose, M₂  =180g/mol

then $∆{\mathrm{T}}_{\mathrm{f}}$ =K_fm

$∆{\mathrm{T}}_{\mathrm{f}}$ =

$\frac{{\mathrm{K}}_{\mathrm{f}}\times {\mathrm{W}}_2\times 1000}{{\mathrm{W}}_1\times {\mathrm{M}}_2} = \frac{13.97\times 5\times 1000}{100\times 180} = 3.88K$

then freezing point of solution = 273.15-3.88
                                              =263.27K
      

Answer:

The vapour pressure of a pure solvent decrease .
when a non-volatile solute is added to the solvent 
this is because on adding the solute a fewer number of water molecules are present at the surface which can evaporate as some of the area is occupied by -non- volatile solute molecules thereby decreasing the vapour pressure of the solution of the glucose in wateris lower than pure water.

Answer;

Let the density of solution = d g/cm³
and volume of solution is = 1L = 1000 cm³

mass of the solution = d xV
                               = (1000d)g

6.90 M solution mean 1L solution contains 6.90 moles of KOH
therefore mass of KOH = 6.90 x 56 = 386.4g

but only 30% of the solutionby mass is KOH
therefore 

30/100 x (1000)d =386.4

and density is 1.288g/ cm³