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When excess lead nitrate solution was ad...

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)

Text Solution

Verified by Experts

The given equation is :

303 g of `PbSO_(4)` is obtained from `Na_(2)SO_(4) = 142 g `
15.15 g of `PbSO_(4)` is obtained from `Na_(2)SO_(4)`
`=142/303xx15.15=7.1`
`therefore` Mass of sodium sulphate = 7.1 g
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