(a) A 10 % solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10 % glucose in water if the freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol⁻¹) (Molar mass of glucose = 180 g mol⁻¹)
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass

OR

(a) 30 g of urea (M = 60 g mol−1 ) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23·8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.

(a) Here $\mathrm{\Delta T}=(273.15-269.15)\mathrm{K}=4\mathrm{K}$
Molar mass of sugar  $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)=12\times 12+22\times 1+11\times 16=342\mathrm{g}{\mathrm{mol}}^{-1}$
10% solution (by mass) of sucrose (cane sugar) in water means 10 g of cane sugar is present in (100 - 10) g = 90 g of water.
Now, number of moles of cane sugar $=\frac{10}{342}=0.0292\mathrm{mol}$
Therefore, molality (m) of the solution, $=\frac{0.0292\times 1000}{90}=0.3244\mathrm{mol}{\mathrm{kg}}^{-1}$
We know that,
$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}.}\mathrm{m}\\ {\mathrm{K}}_{\mathrm{f}}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{\mathrm{m}}\\ =\frac{4}{0.3244}=12.33\mathrm{Kkg}{\mathrm{mol}}^{-1}\end{array}$
Molar mass of glucose $\left(\mathrm{C}\left({\mathrm{H}}_{12}{\mathrm{O}}_6\right)=6\times 12+12\times 1+6\times 16=180\mathrm{g}{\mathrm{mol}}^{-1}\right.$
10% glucose in water means 10g of glucose is present in (100 - 10) g = 90 g of water.
∴ Number of mole of glucose $=\frac{10}{180}\mathrm{mol}=0.0555\mathrm{mol}$
Therefore, molality (m) of the solution $=\frac{0.0555\times 1000}{\left(\mathrm{K}\right)}=0.6166\mathrm{mol}{\mathrm{kg}}^{-1}$
Applying the relation, 
$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\cdot \mathrm{m}\\ =12.33\mathrm{Kkg}{\mathrm{mol}}^{-1}\times 0.6166\mathrm{mol}{\mathrm{kg}}^{-1}\\ =7.60\mathrm{K}\text{ (approximately) }\end{array}$
Hence, the freezing point of 10% glucose solution is (273.15- 7.60) K
K=265.55K

(b) (i) Molality:
The number of moles of solute present in 1000 g of solvent is called molality.
It is denoted by m.

(ii) Abnormal molar mass: When the molar mass is higher or lower than actual molar mass due to association or dissociation of molecules, it is called abnormal molar mass.

OR

(a) It is given that vapour pressure of water, 𝑃°₁ = 23.8 mm of Hg
Weight of water taken, w₁ =846 g
Weight of urea taken, w₂ = 30 g
Molecular weight of water M₁ = 18 g mol^-1
Molecular weight of urea, M₂ = 60 g mol^-1
Now, we have to calculate vapour pressure of water in the solution we take vapour pressure as P₁.
Now from Raoult's law, we have:
$\begin{array}{l}\frac{{\mathrm{P}}_1^0-{\mathrm{P}}_0}{{\mathrm{P}}_1^0}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1-{\mathrm{n}}_2}\\ \frac{{\mathrm{P}}_1^0-{\mathrm{P}}_0}{{\mathrm{P}}_1^0}=\frac{\frac{{\mathrm{m}}_2}{{\mathrm{M}}_2}}{\frac{{\mathrm{m}}_1}{{\mathrm{M}}_1}-\frac{{\mathrm{m}}_2}{{\mathrm{M}}_2}}\\ \Rightarrow \frac{23.8-{\mathrm{P}}_1}{23.8}=0.0105\\ \Rightarrow {\mathrm{P}}_1=23.5501\mathrm{mm}\text{ of }\mathrm{Hg}\end{array}$
Hence, the vapour pressure of water in the given solution is 23.5501 mm of Hg and its relative lowering is 0.0105.

(b)