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On mixing 45.0 mL of 0.25 M lead nitrate...

On mixing `45.0 mL` of `0.25 M` lead nitrate solution with `25.0 mL` of `0.10 M` chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molar concentration of the species left behind solution. Assume the lead sulphate is completely insoluble.

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The correct Answer is:
`7.14 xx 10^(-2) M`
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