The detailed solutions to the 10 simple numericals from the chapter "Solutions" (Class 12 Chemistry)

 The detailed solutions to the 10 simple numericals from the chapter "Solutions" (Class 12 Chemistry):

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1. Calculate the molarity of a solution prepared by dissolving 5.8 g of NaCl in 500 mL of solution.

Given:
Mass of NaCl = 5.8 g
Volume of solution = 500 mL = 0.5 L
Molar mass of NaCl = 58.5 g/mol

Solution:

$$\text{Moles of NaCl} = \frac{5.8}{58.5} = 0.1 \, \text{mol}$$
$$\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{0.1}{0.5} = 0.2 \, \text{M}$$

✅ Answer: 0.2 M

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2. Find the molality of a solution containing 36 g of glucose (C₆H₁₂O₆) in 250 g of water.

Given:
Mass of glucose = 36 g
Molar mass = 180 g/mol
Mass of water = 250 g = 0.25 kg

Solution:

$$\text{Moles of glucose} = \frac{36}{180} = 0.2 \, \text{mol}$$
$$\text{Molality (m)} = \frac{0.2}{0.25} = 0.8 \, \text{mol/kg}$$

✅ Answer: 0.8 mol/kg

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3. Calculate the mass of urea to lower freezing point by 0.93°C in 100 g water.

Given:
ΔTf = 0.93°C
Kf = 1.86 K·kg/mol
Mass of water = 100 g = 0.1 kg
Molar mass of urea = 60 g/mol

Solution:

$$\Delta T_f = K_f \times m \Rightarrow m = \frac{0.93}{1.86} = 0.5 \, \text{mol/kg}$$
$$\text{Moles of urea} = 0.5 \times 0.1 = 0.05 \, \text{mol}$$
$$\text{Mass of urea} = 0.05 \times 60 = 3.0 \, \text{g}$$

✅ Answer: 3.0 g

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4. Determine the boiling point of a solution with 1 mol solute in 1 kg water.

Given:
n = 1 mol
m = 1 mol/kg
Kb = 0.52 K·kg/mol
Normal boiling point = 373 K

Solution:

$$\Delta T_b = K_b \cdot m = 0.52 \cdot 1 = 0.52\, \text{K}$$
$$\text{New boiling point} = 373 + 0.52 = 373.52\, \text{K}$$

✅ Answer: 373.52 K

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5. A 10 g solute in 200 g water boils at 100.26°C. Find molar mass.

Given:
ΔTb = 0.26°C
Kb = 0.52
Mass of water = 200 g = 0.2 kg
Mass of solute = 10 g

Solution:

$$m = \frac{\Delta T_b}{K_b} = \frac{0.26}{0.52} = 0.5 \, \text{mol/kg}$$
$$\text{Moles} = m \times \text{mass of solvent} = 0.5 \times 0.2 = 0.1 \, \text{mol}$$
$$\text{Molar mass} = \frac{\text{mass of solute}}{\text{moles}} = \frac{10}{0.1} = 100 \, \text{g/mol}$$

✅ Answer: 100 g/mol

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6. Calculate osmotic pressure of 0.1 M sucrose at 27°C.

Given:
C = 0.1 M, T = 27°C = 300 K
R = 0.0821 L·atm/mol·K

Solution:

$$\pi = CRT = 0.1 \times 0.0821 \times 300 = 2.463 \, \text{atm}$$

✅ Answer: 2.463 atm

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7. Vapour pressure of solution = 95 mm Hg; pure solvent = 100 mm Hg. Find mole fraction of solute.

Solution:

$$\frac{P^0 - P}{P^0} = X_{\text{solute}} = \frac{100 - 95}{100} = 0.05$$

✅ Answer: Mole fraction of solute = 0.05

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8. Find mass of NaCl to lower freezing point of 500 g water to -1.86°C.

Given:
ΔTf = 1.86°C, Kf = 1.86, i = 2
Mass of water = 0.5 kg
Molar mass of NaCl = 58.5 g/mol

Solution:

$$m = \frac{\Delta T_f}{i \cdot K_f} = \frac{1.86}{2 \cdot 1.86} = 0.5 \, \text{mol/kg}$$
$$\text{Moles of NaCl} = 0.5 \times 0.5 = 0.25 \, \text{mol}$$
$$\text{Mass} = 0.25 \times 58.5 = 14.625 \, \text{g}$$

✅ Answer: 14.625 g

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9. Calculate mole fraction of ethanol in 46 g ethanol + 180 g water.

Molar masses:
Ethanol = 46 g/mol, Water = 18 g/mol

Solution:

$$\text{Moles of ethanol} = \frac{46}{46} = 1 \, \text{mol}, \quad \text{Moles of water} = \frac{180}{18} = 10 \, \text{mol}$$
$$X_{\text{ethanol}} = \frac{1}{1+10} = \frac{1}{11} \approx 0.091$$

✅ Answer: 0.091

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10. A 5% (w/v) solution exerts osmotic pressure of 4.1 atm at 27°C. Find molar mass.

Given:
5% w/v → 5 g solute in 100 mL → 0.1 L
T = 300 K, R = 0.0821 L·atm/mol·K
π = 4.1 atm

Solution:

$$\pi = \frac{w}{M \cdot V} \cdot R T$$
$$4.1 = \frac{5}{M \cdot 0.1} \cdot 0.0821 \cdot 300$$
$$4.1 = \frac{5 \cdot 24.63}{0.1 \cdot M} = \frac{123.15}{0.1 \cdot M} \Rightarrow 4.1 \cdot 0.1 \cdot M = 123.15 \Rightarrow M = \frac{123.15}{0.41} \approx 300.37$$

✅ Answer: Approx. 300.4 g/mol