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Standard XII

Chemistry

Laws of Mass Conservation

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.

525.0

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525

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

525.00

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Solution

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

$mmol of C r_{2} (S O_{4})_{3} = 20 \times 0.12 = 2.4 mmol$
$P b (N O_{3})_{2} = 35 \times 0.15 = 5.25 mmol$
$3 mmol P b (N O_{3})_{2} reacts with 1 mmol C r_{2} (S O_{4})_{3}$
$5.25 mmol P b (N O_{3})_{2} reacts with 1.75 mmol C r_{2} (S O_{4})_{3}$
Limiting reagent is lead nitrate
So, $5.25 mmol$ 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.

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.