When excess lead nitrate solution was added to a solution of sodium sulphate, 15.15 g of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
`Na_(2)SO_(4) +Pb(NO_(3))_(2) to PbSO_(4)+2NaNO_(3)`
(O=16,Na=23,S=32,Pb=207)

Explanation

To find the mass of sodium sulfate (Na₂SO₄) present in the original solution when 15.15 g of lead sulfate (PbSO₄) was precipitated, we can follow these steps:

### Step 1: Calculate the molar mass of sodium sulfate (Na₂SO₄)
- Sodium (Na) has a molar mass of 23 g/mol. Since there are 2 sodium atoms:
\[ 2 \times 23 = 46 \text{ g/mol} \]
- Sulfur (S) has a molar mass of 32 g/mol.
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