A sample of 10 g of a mixture of sodium chloride and anhydrous sodium sulphate of dissolved in water. When an excess of barium chloride solution is added, 6.99 g of barium sulphate is precipitated according to the equation: `Na_(2)SO_(4)+BaCl_(2)to BaSO_(4)+2NACl`. Calculate the percentage of sodium sulphate in the original mixture. (O = 16, Na = 23,S = 32, Ba = 137)

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the molar masses of sodium sulfate and barium sulfate.

1. **Molar mass of Sodium Sulfate (Na₂SO₄)**:
- Sodium (Na) = 23 g/mol, and there are 2 sodium atoms: \(2 \times 23 = 46 \, \text{g/mol}\)
- Sulfur (S) = 32 g/mol: \(1 \times 32 = 32 \, \text{g/mol}\)
- Oxygen (O) = 16 g/mol, and there are 4 oxygen atoms: \(4 \times 16 = 64 \, \text{g/mol}\)
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