i = \({observed\ property\over Theoritical\ property\ (calculated)}\)
Given ΔT_f = 0.680 K
m = 0.2 m
ΔT_f (observed) = 0.680 K
ΔT_f (calculated) = K_f m
= 1.86 K Kg mol^–1 × 0.2 mol Kg^–1
= 0.372 K
i = \({(\Delta T_f)\ observed\over (\Delta T_f)\ calculated}={0.680\ K\over 0.372\ K}=1.82\)

Osmotic pressure of urea solution (\(\pi_1\)) = CRT
\(={W_2\over M_2V}RT\)
\(={6\over 60\times 1}\times RT\)
Osmotic pressure of glucose solution \((\pi_2)={W_2\over 180\times 1}\times RT\) For isotonic solution,
\(\pi_1=\pi_2\)
\({6\over 60}RT={W_2\over 180}RT\)
\(\Rightarrow W_2={6\over 60}\times 180\)
\(W_2=18\ g\)

No.of moles ; n = \(\frac{\mathrm{m}}{\mathrm{M}}=\frac{5.6}{56}=0.1 \mathrm{~mol}\)
(i) V = 500 ml = \(\frac{500}{1000}=0.5 \mathrm{~L} \)
Molarity = \(\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1}{0.5}=0.2 \mathrm{M} \)
(ii) V = IL 
Molarity = \(\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1}{1}=0.1 \mathrm{M} \)

Molar mass of solute
\(={1000\times K_f\times mass\ of\ NaCl\over \Delta T_f\times mass\ of\ solvent}\)
\(={1000\times 1.86\times 1\over 0.24\times 200}\)
= 38.75 g mol^-1
= 38.75 g mol
Theoretical molar mass of NaCl is 
i = \({Theoretical\ molar\ mass\over Experimental\ molar\ mass}={58.5\over 38.75}=1.50\)

W₂ = 2g, W₁ = 75 g
ΔT_f = 0.2 K , K_f = 5.12 K Kg mol^–1
M₂ = ?
\(M_2={K_f\times W_2\times 1000\over \Delta T_f\times W_1}={5.12\times 2\times 1000\over 0.2\times 75}\)
= 682.66 g mol^-1.

Weight of solute (W₂) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W₁ = 100 - 20 = 80 g
ΔT_f = K_f m
\(={K_f\times W_2\times 1000\over M_2\times W_1}\)
\(={1.86\times 20\times 1000\over 62\times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 - 7.5K) = 265.5 K

\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
In a 2 % solution weight of the solute is 2g and solvent is 98g
ΔP = P_A - P_solution = 1.013 - 1.004 bar = 0.009 bar
\(M_B={P_A^o\times W_B\times M_A\over \Delta P\times W_A}\)
M_B = 2 x 18 x 1.013/(98 x 0.009)
= 41.3 g mol^-1.

\(P_A^o\) = 10 torr, P_solution = 9 torr
W_A = 20 g  W_B = 1 g
M_A = 200 g mol^-1  M_B = ?
\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
\({10-9\over 10}={1\times 200\over M_B\times 20}\)
\(M_B={200\over 20}\times 10=100\ g\ mol^{-1}\)

(a) mass of 1.5 moles of CoCl₂ = 1.5 x 129.9
= 194.85 g
194.85 g anhydrons cobalt chloride is dissolved in water and the solution is make up to one litre in a standard flask.
(b) \(6\%{V\over V}\)  aqeous solution contains 6g of methanol in 100 ml solution.
\(\therefore\) To prepare 500 ml of  \(6\%{V\over V}\) solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

\(\Delta T_f=0.093^oC=0.093K\)
m = ?
K_f = 1.86K Kg mol^-1
\(\Delta T_f=K_f.m\)
\(\therefore m={\Delta T_f\over K_f}\)
\(={0.093K\over 1.86\ K\ Kg\ mol^{-1}}\)
= 0.05 mol Kg^-1
= 0.05 m.