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Chemistry Question on Expressing Concentration of Solutions

The 25 mL25 \,mL of a 0.15 M0.15 \, M solution of lead nitrate, Pb(NO3)2Pb \left( NO _{3}\right)_{2} reacts with all of the aluminium sulphate, Al2(SO4)3Al _{2}\left( SO _{4}\right)_{3}, present in 20 mL20 \,mL of a solution. What is the molar concentration of the Al2(SO4)3Al _{2}\left( SO _{4}\right)_{3} ? 3Pb(NO3)2(aq)+Al2(SO4)3(aq)⟶3PbSO4(s)+2Al(NO3)3(aq)3 Pb \left( NO _{3}\right)_{2}( aq )+ Al _{2}\left( SO _{4}\right)_{3}( aq ) \longrightarrow 3 PbSO _{4}( s )+2 Al \left( NO _{3}\right)_{3}( aq )

A

6.25×10−2 M6.25 \times 10^{-2} \, M

B

2.421×10−2 M2.421 \times 10^{-2} \,M

C

0.1875 M

D

None of these

Answer

6.25×10−2 M6.25 \times 10^{-2} \, M

Explanation

Solution

The correct option is (A): 6.25×10−2 M6.25 \times 10^{-2} \, M.
Molar mass of Pb(NO3)2=25×0⋅15Pb \left( NO _{3}\right)_{2}=25 \times 0 \cdot 15
=3.75 m=3.75\, m. moles
Molar mass of Al2(SO4)3Al _{2}\left( SO _{4}\right)_{3}
=13×3⋅75=M×20=\frac{1}{3} \times 3 \cdot 75= M \times 20
M=0.0625=6.25×10−2MM=0.0625=6.25 \times 10^{-2} M