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First Order Reaction

25 mL of 0.15...

Question

25 mL of 0.15 M Pb $(N O_{3})_{2}$ reacts completely with 20 mL of $A l_{2} (S O_{4})_{3}$ . The molar concentration of $A l_{2} (S O_{4})_{3}$ will be: . . .
$3 P b (N O_{3})_{2} (a q) + A l_{2} (S O_{4})_{3} (a q .) \to 3 P b S O_{4} (s) + 2 A l (N O_{3})_{3} (a q .)$

0.0625 M

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0.0242 M

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0.1875 M

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0.2346 M

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Solution

The correct option is A 0.0625 M
$\frac{M_{1} V_{1}}{n_{1}} P b (N O_{3})_{2} = \frac{M_{2} V_{2}}{n_{2}} A l_{2} (S O_{4}) 3$

$\frac{0.15 \times 25}{3} = \frac{M_{2} \times 20}{1}$

$M_{2} = 0.0625$

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