Calculate the amount of each in the following solutions -
(i) 150 ml of `N/7H_2SO_4`
(ii) 250 ml of 0.2M `NaHCO_3`
(iii) 400 ml of `N/10Na_2CO_3`
(iv) 1056 g of 1 m KOH

Options

A. 52g, 2.12g, 4.2g, 1.05g
B. 1.05g, 4.2g, 2.12g, 56g
C. 1.05g, 2.12g, 52g, 4.2g
D. 4.2g, 2.12g, 1.05g, 52g

Explanation

To solve the problem of calculating the amount of solute in the given solutions, we will use the formulas for normality and molarity. Let's break it down step by step for each part.

### (i) 150 ml of `N/7 H2SO4`

1. **Identify the given data**:
- Volume (V) = 150 ml
- Normality (N) = 1/7

2. **Calculate the equivalent mass of H2SO4**:
- Molecular mass of H2SO4 = 2(1) + 32 + 4(16) = 98 g/mol
- Basicity of H2SO4 = 2 (since it can donate 2 H⁺ ions)
- Equivalent mass = Molecular mass / Basicity = 98 g/mol / 2 = 49 g/equiv

3. **Use the normality formula**:
\[ N = \frac{W}{\text{Equivalent mass} \times \text{Volume in L}} \]
Rearranging gives:
\[ W = N \times \text{Equivalent mass} \times \text{Volume in L} \]
- Convert volume to liters: 150 ml = 0.150 L
- Substitute values:
\[ W = \frac{1}{7} \times 49 \times 0.150 \]
\[ W = 1.05 \text{ grams} \]

### (ii) 250 ml of 0.2M `NaHCO3`

1. **Identify the given data**:
- Volume (V) = 250 ml
- Molarity (M) = 0.2 M

2. **Calculate the molecular mass of NaHCO3**:
- Molecular mass = 23 (Na) + 1 (H) + 12 (C) + 3(16) (O) = 84 g/mol

3. **Use the molarity formula**:
\[ M = \frac{W}{\text{Molecular mass} \times \text{Volume in L}} \]
Rearranging gives:
\[ W = M \times \text{Molecular mass} \times \text{Volume in L} \]
- Convert volume to liters: 250 ml = 0.250 L
- Substitute values:
\[ W = 0.2 \times 84 \times 0.250 \]
\[ W = 4.2 \text{ grams} \]

### (iii) 400 ml of `N/10 Na2CO3`

1. **Identify the given data**:
- Volume (V) = 400 ml
- Normality (N) = 1/10

2. **Calculate the equivalent mass of Na2CO3**:
- Molecular mass of Na2CO3 = 2(23) + 12 + 3(16) = 106 g/mol
- Basicity of Na2CO3 = 2 (since it can accept 2 H⁺ ions)
- Equivalent mass = Molecular mass / Basicity = 106 g/mol / 2 = 53 g/equiv

3. **Use the normality formula**:
\[ N = \frac{W}{\text{Equivalent mass} \times \text{Volume in L}} \]
Rearranging gives:
\[ W = N \times \text{Equivalent mass} \times \text{Volume in L} \]
- Convert volume to liters: 400 ml = 0.400 L
- Substitute values:
\[ W = \frac{1}{10} \times 53 \times 0.400 \]
\[ W = 2.12 \text{ grams} \]

### (iv) 1056 g of 1 m KOH

1. **Identify the given data**:
- Total weight of solution = 1056 g
- Molality (m) = 1 m

2. **Understand the definition of molality**:
- 1 molal means 1 mole of solute in 1 kg of solvent.

3. **Calculate the weight of the solvent**:
- Since 1 molal means 1 kg of solvent, the weight of the solvent = 1000 g.
- Therefore, the weight of the solute (KOH) = Total weight - Weight of solvent = 1056 g - 1000 g = 56 g.

### Final Results:
- (i) 1.05 g of H2SO4
- (ii) 4.2 g of NaHCO3
- (iii) 2.12 g of Na2CO3
- (iv) 56 g of KOH