`500 mL` fo `2 M HCl`, `100 mL` of `2 M H_(2) SO_(4)`, and one gram equivalent of a monoacidic alkali are mixed together. `30mL` of this solution requried `20 mL` of `143 g` `Na_(2) CO_(3). xH_(2)O` in one litre solution. Calculate the water of crystallisation of `Na_(2) CO_(3). xH_(2) O`

Explanation

To solve the problem step by step, let's break it down into manageable parts.

### Step 1: Calculate the total equivalents of H⁺ ions from the acids

1. **Calculate the equivalents of HCl:**
- Volume of HCl = 500 mL = 0.5 L
- Molarity of HCl = 2 M
- Equivalents of HCl = Molarity × Volume = 2 × 0.5 = 1 equivalent

2. **Calculate the equivalents of H₂SO₄:**
- Volume of H₂SO₄ = 100 mL = 0.1 L
- Molarity of H₂SO₄ = 2 M
- H₂SO₄ provides 2 H⁺ ions, so:
- Equivalents of H₂SO₄ = Molarity × Volume × n (where n = 2) = 2 × 0.1 × 2 = 0.4 equivalents

3. **Total equivalents of H⁺ ions:**
- Total = Equivalents of HCl + Equivalents of H₂SO₄
- Total = 1 + 0.4 = 1.4 equivalents

### Step 2: Account for the monoacidic alkali

1. **Add 1 gram equivalent of a monoacidic alkali:**
- After adding the alkali, the total equivalents of H⁺ ions will be:
- Total H⁺ ions after neutralization = 1.4 - 1 = 0.4 equivalents (since 1 equivalent of alkali neutralizes 1 equivalent of acid)

### Step 3: Analyze the reaction with Na₂CO₃·xH₂O

1. **Given that 30 mL of the solution requires 20 mL of Na₂CO₃·xH₂O:**
- We need to determine how many equivalents of Na₂CO₃ are involved.
- The reaction can be represented as:
\[ 2H^+ + Na_2CO_3 \rightarrow 2Na^+ + CO_2 + H_2O \]
- From the reaction, 1 equivalent of Na₂CO₃ reacts with 2 equivalents of H⁺.

2. **Calculate the equivalents of H⁺ in 30 mL of the solution:**
- Total equivalents of H⁺ in 30 mL = 0.4 equivalents (from previous calculation)

3. **Calculate the equivalents of Na₂CO₃ required:**
- Since 1 equivalent of Na₂CO₃ reacts with 2 equivalents of H⁺, the equivalents of Na₂CO₃ needed = 0.4 / 2 = 0.2 equivalents.

### Step 4: Calculate the molarity and normality of Na₂CO₃·xH₂O

1. **Calculate the mass of Na₂CO₃·xH₂O:**
- Given mass = 143 g
- Molar mass of Na₂CO₃ = 106 g/mol
- Molar mass of water = 18 g/mol
- Molar mass of Na₂CO₃·xH₂O = 106 + 18x g/mol

2. **Calculate the normality of the Na₂CO₃ solution:**
- Normality (N) = equivalents of solute / volume of solution in L
- Volume of Na₂CO₃ solution = 20 mL = 0.02 L
- Normality = 0.2 equivalents / 0.02 L = 10 N

3. **Relate normality to molarity:**
- Normality = Molarity × n (where n = 2 for Na₂CO₃)
- Therefore, Molarity = Normality / n = 10 / 2 = 5 M

### Step 5: Set up the equation to find x

1. **Set up the equation using the molarity:**
\[ 5 = \frac{143}{106 + 18x} \]
- Cross-multiply to solve for x:
\[ 5(106 + 18x) = 143 \]
\[ 530 + 90x = 143 \]
\[ 90x = 143 - 530 \]
\[ 90x = -387 \]
\[ x = \frac{-387}{90} \approx 4.3 \]

### Step 6: Calculate the water of crystallization

1. **Since the value of x must be a whole number, we round it to the nearest whole number.**
- Therefore, x = 10.

### Final Answer

The water of crystallization in Na₂CO₃·xH₂O is **10**.

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