No. of, moles of glucose ; n₂ = \(\frac{\mathrm{m}}{\mathrm{M}}=\frac{2.82}{180}=0.016 \mathrm{~mol} \)
No. of. moles of water ; n₁ = \(\frac{30}{18}=1.67 \mathrm{~mol} \)
\(\mathrm{X}_{1}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1.67}{1.67+0.016}=0.99\)
\(\mathrm{X}_{1}+\mathrm{X}_{2}=1 \)
\(\therefore \mathrm{X}_{2}=1-\mathrm{X}_{1}=1-0.99=0.01 \).

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

In 10^–4 M K₂SO₄ solution, there are 10^–4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K⁺ and 1So₄^2–)
1 mole of K₂SO₄ contains 3 x 6.023 x 10²³ ions
10^–4 mole of K₂SO ₄ contains 3 x 6.023 x 1023 x 10^–4 ions = 18.069 x 10¹⁹.

\(\Delta T_f=K_f\ m\)
ie \(\Delta T_f\alpha\ m\)
\(m_{CH_3-OH}={({10\over 32})\over 0.1}=3.125\ m\)
\(m_{C_2H_5-OH}={({20\over 46})\over 0.2}=2.174\ m\)
\(\therefore\) depression in freezing point is more in methanol solution and it will have lower freezing point.

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 = \({Theoretical\ molar\ mass\over Experimental\ molar\ mass}={58.5\over 38.75}=1.50\)

ΔT_b = K_b m
= K_b x W₂ x 1000 / M₂ x W₁
M₂ = K_b x W₂ x 1000 / ΔT_b x W₁
= 7.5 x 0.75 x 1000 / 0.15 x 200
= 187.5 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 = ?
\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
\({10-9\over 10}={1\times 200\over M_B\times 20}\)
\(M_B={200\over 20}\times 10=100\ g\ mol^{-1}\)

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.

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.