Question

When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, x×10−5 moles of lead sulphate are precipitated out.The value of x Rounded off to the nearest integer is

Answer 1

3Pb(NO3)2+Cr2(SO4)3→3PbSO4+2Cr(NO3)3
mmol of Cr2(SO4)3=20×0.12=2.4 mmol
Pb(NO3)2=35×0.15=5.25 mmol
3 mmol Pb(NO3)2reacts with 1 mmol Cr2(SO4)3
5.25 mmol Pb(NO3)2reacts with 1.75 mmol Cr2(SO4)3
Limiting reagent is lead nitrate
So, 5.25mmol of lead sulphate are precipitated.
Moles of PbSO4 formed=5.25×10−3=525×10−5
value of x is 525