Question

25mL of 0.15M Pb `(NO_(3))_(2)` reacts completely with 20 mL of `Al_(2)(SO_(3))_(2)`. The molar concentration of `Al_(2)(SO_(4))_(3)` will be:
`3Pb(NO_(3))_(2)(aq.)+Al_(2)(SO_(4))_(3)(aq.)to3PbSO_(3)(s)+2Al(NO_(3))_(3)(aq.)`

Answer 1

Correct Answer - a
`(M_(1)V_(1))/(n_(1))Pb(NO_(3))_(2)=(M_(2)V_(2))/(n_(2))Al_(2)(SO_(4))_(3)`
`(0.15xx25)/(3)=(M_(2)xx20)/(1)`
`M_(2)`=0.0625