The 25 mL of a 0.15 M solution of lead nitrate, Pb ( NO 3)2 reacts with all of the aluminium sulphate, Al 2( SO 4)3, present in 20 mL of a solution. What is the molar concentration of the Al 2( SO 4)3 ? 3 Pb ( NO 3)2( aq )+ Al 2( SO 4)3( aq ) longrightarrow 3 PbSO 4( s )+2 Al ( NO 3)3( aq )

Question

The 25 mL of a 0.15 M solution of lead nitrate, Pb ( NO 3)2 reacts with all of the aluminium sulphate, Al 2( SO 4)3, present in 20 mL of a solution. What is the molar concentration of the Al 2( SO 4)3 ? 3 Pb ( NO 3)2( aq )+ Al 2( SO 4)3( aq ) longrightarrow 3 PbSO 4( s )+2 Al ( NO 3)3( aq ) (A) 6.25 × 10-2 M  (B) 2.421 × 10-2 M  (C) 0.1875 M (D) None of these

Tardigrade Admin · Accepted Answer

Molar mass of Pb ( NO 3)2=25 × 0 ⋅ 15 =3.75 m. moles Molar mass of Al 2( SO 4)3 =(1/3) × 3 ⋅ 75= M × 20 M=0.0625=6.25 × 10-2 M

$6.25 \times 1 0^{- 2} M$

$2.421 \times 1 0^{- 2} M$

0.1875 M

None of these

Molar mass of $P b (N O_{3})_{2} = 25 \times 0 \cdot 15$
$= 3.75 m$ . moles
Molar mass of $A l_{2} (S O_{4})_{3}$
$= \frac{1}{3} \times 3 \cdot 75 = M \times 20$
$M = 0.0625 = 6.25 \times 1 0^{- 2} M$

Q. The 25mL of a 0.15M solution of lead nitrate, Pb(NO3​)2​ reacts with all of the aluminium sulphate, Al2​(SO4​)3​, present in 20mL of a solution. What is the molar concentration of the Al2​(SO4​)3​ ? 3Pb(NO3​)2​(aq)+Al2​(SO4​)3​(aq)⟶3PbSO4​(s)+2Al(NO3​)3​(aq)

6.25×10−2M

2.421×10−2M

0.1875 M

None of these

Molar mass of Pb(NO3​)2​=25×0⋅15 =3.75m. moles Molar mass of Al2​(SO4​)3​ =31​×3⋅75=M×20 M=0.0625=6.25×10−2M

$6.25 \times 1 0^{- 2} M$

$2.421 \times 1 0^{- 2} M$

Molar mass of $P b (N O_{3})_{2} = 25 \times 0 \cdot 15$
$= 3.75 m$ . moles
Molar mass of $A l_{2} (S O_{4})_{3}$
$= \frac{1}{3} \times 3 \cdot 75 = M \times 20$
$M = 0.0625 = 6.25 \times 1 0^{- 2} M$