Solutions Class 12 Chemistry NCERT Solutions

Step-by-step solution

1. 1Total mass of solution = 22 + 122 = 144 g. Mass percentage of each component:
2. 2
   $$\%\text{ benzene}=\frac{22}{144}\times 100=15.28\%$$
3. 3
   $$\%{\text{ CCl}}_4=\frac{122}{144}\times 100=84.72\%$$

Step-by-step solution

1. 1Take 100 g of solution: 30 g benzene $(M=78)$ and 70 g CCl₄ $(M=154)$. Moles of each:
2. 2
   $$n_{\text{benzene}}=\frac{30}{78}=0.3846$$
3. 3
   $$n_{{\text{CCl}}_4}=\frac{70}{154}=0.4545$$
4. 4
   $$x_{\text{benzene}}=\frac{0.3846}{0.3846+0.4545}=\frac{0.3846}{0.8392}\approx 0.46$$

Step-by-step solution

1. 1Moles of NaCl, then molarity (volume in litres):
2. 2
   $$n=\frac{5.85}{58.5}=0.1\text{mol}$$
3. 3
   $$M=\frac{n}{V_{(\text{L})}}=\frac{0.1}{0.500}=0.2{\text{mol L}}^{-1}$$

Step-by-step solution

1. 1Relative lowering of vapour pressure equals the solute mole fraction:
2. 2
   $$\frac{p^{\circ }-p}{p^{\circ }}=x_{\text{solute}}$$
3. 3
   $$x_{\text{solute}}=\frac{30.8-29.0}{30.8}=\frac{1.8}{30.8}\approx 0.058$$

Step-by-step solution

1. 1Molality is the moles of solute dissolved per kilogram of solvent:
2. 2
   $$m=\frac{n_{\text{solute}}}{\text{kg solvent}}$$
3. 3For a solution of solvent A (Mₐ kg mol⁻¹) and solute B, $m=\frac{1000x_B}{M_a(1-x_B)}{\text{mol kg}}^{-1}$ — the direct link between molality and mole fraction.

Step-by-step solution

1. 1Moles of glucose and molality:
2. 2
   $$n=\frac{6}{180}=0.0333\text{mol}$$
3. 3
   $$m=\frac{0.0333}{0.250}=0.1333{\text{mol kg}}^{-1}$$
4. 4Elevation of boiling point (non-electrolyte, i = 1):
5. 5
   $$\Delta T_b=K_b\times m=0.52\times 0.1333\approx 0.069\text{K}$$

Step-by-step solution

1. 1NaCl dissociates into 2 ions (Na⁺, Cl⁻) so the van't Hoff factor i = 2:
2. 2
   $$n=\frac{0.585}{58.5}=0.01\text{mol},m=\frac{0.01}{0.100}=0.1{\text{mol kg}}^{-1}$$
3. 3
   $$\Delta T_f=i{K}_{f}m=2\times 1.86\times 0.1\approx 0.372\text{K}$$