Chapter 1: Solutions

Ans: Given that,

Lowering in freezing point, $\Delta {{T}_{f}}$  = 1.5${}^\circ $ C

Cryoscopic constant, ${{K}_{f}}=3.9Kg/mol$

Mass of acetic acid, ${{W}_{1}}$ = 75 g

Molar mass of acetic acid, ${{M}_{1}}$ = 60 g/mol

Molar mass of ascorbic acid, ${{M}_{2}}$ = 176 g/mol

Mass of ascorbic acid, ${{W}_{2}}=?$

The lowering in freezing point is given as;

 \[\Delta {{T}_{f}}={{K}_{f}}\times m\]

\[\Delta {{T}_{f}}={{K}_{f}}\times \dfrac{{{n}_{2}}}{{{W}_{1}}}={{K}_{f}}\times \dfrac{{{W}_{2}}}{{{M}_{2}}}\times \dfrac{1000}{{{W}_{1}}}\]

\[{{W}_{2}}=\dfrac{\Delta {{T}_{f}}\times {{M}_{2}}\times {{W}_{1}}}{{{K}_{f}}\times 1000}=5.077g\]

$\therefore $ Mass of ascorbic acid is 5.077 g.

Ans: Given that,

Mass of polymer, W = 1 g

Molar mass of polymer, M = 185000 g/mol

Volume of water, V = 450 ml = 0.45 L

Temperature, T = 37${}^\circ $ C = 310 K

Gas constant, R = $8.314kPa.L.{{K}^{-1}}mo{{l}^{-1}}=8.314\times {{10}^{3}}Pa.L.{{K}^{-1}}mo{{l}^{-1}}$ 

Now, using the formula for osmotic pressure;

\[\pi =CRT=\dfrac{n}{V}RT=\dfrac{W}{M}\times \dfrac{1}{V}\times RT\] 

\[\pi =30.96Pa\] 

$\, therefore, $ Osmotic pressure is 30.96 Pa.