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 = \(\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1}{1}=0.1 \mathrm{M} \)

Let the volume of 4M HCl required to prepare 500 mL of 2.5 MHCl = x mL
Therefore, the required volume of 2M HCl = (500 - x) mL
We know from the equation
C₁V₁+ C₂V₂ = C₃V₃
(4x) + 2(500 - x) = 2.5 x 500
4x + 1000 - 2x = 1250
2x = 1250 - 1000
\(x={250\over 2}\)
= 125 ml
Hence, volume of 4M HCl required = 125 mL
Volume of 2M HCl required = (500 - 125) mL = 375 mL.

P_solute  = K_H X_{solute in solution}
At pressure 1.5 atm,
p₁ = K_H x_{1       ------}(1)
At pressure 6.0 atm,
p₂ = K_H x₂        -----(2)
Dividing equation (1) by (2)
From equation p₁/p₂ = x₁/x₂
1.5/6.0 = 0.24/x₂
Therefore x₂ = 0.24 x 6.0/1.5 = 0.96 g/L.

\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
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.

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