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 JEE MAIN 2021

3Pb(NO_3)_2+Cr_2(SO_4)_3→ 3PbSO_4+2Cr(NO_3)_3 mmol of Cr_2(SO_4)_3=20× 0.12=2.4 mmol Pb(NO_3)_2=35× 0.15=5.25 mmol 3 mmol Pb(NO_3)_2reacts with 1 mmol Cr_2(SO ...

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Byju's Answer

Standard X

Chemistry

Mole Concept

When 35 mL of...

Question

When $35 mL$ of $0.15 M$ lead nitrate solution is mixed with $20 mL$ of $0.12 M$ chromic sulphate solution, $x \times 10^{- 5}$ moles of lead sulphate are precipitated out.The value of $x$ Rounded off to the nearest integer is

 JEE MAIN 2021

525.0

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525.00

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525

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Solution

$3 P b (N O_{3})_{2} + C r_{2} (S O_{4})_{3} \to 3 P b S O_{4} + 2 C r (N O_{3})_{3}$

$mmol of C r_{2} (S O_{4})_{3} = 20 \times 0.12 = 2.4 m m o l$
$P b (N O_{3})_{2} = 35 \times 0.15 = 5.25 m m o l$
$3 m m o l P b (N O_{3})_{2} reacts with 1 m m o l C r_{2} (S O_{4})_{3}$
$5.25 m m o l P b (N O_{3})_{2} reacts with 1.75 m m o l C r_{2} (S O_{4})_{3}$
Limiting reagent is lead nitrate
So, 5.25mmol of lead sulphate are precipitated.
$Moles of P b S O_{4} formed = 5.25 \times 10^{- 3} = 525 \times 10^{- 5}$
value of x is 525

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When 35mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12M chromic sulphate solution, x×10−5 moles of lead sulphate are precipitated out.The value of x Rounded off to the nearest integer is  JEE MAIN 2021