What minimum volume of 0.2 M $K I (a q)$ solution is required to completely precipitate all of the lead in 150 ml $0.1 M P b (N O_{3})_{2}$ solution? $2 K I + P b (N O_{3})_{2} (a q) \to 2 K N O_{3} (a q) + P b I_{2} (s)$

300 mL

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75 mL

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600 mL

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150 mL

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Solution

The correct option is D 150 mL
From the balanced chemical equation
1 mole of $P b (N O_{3})_{2}$ requires 2 mole of $K I$ .
Number of mmoles of $P b (N O_{3})_{2}$ $= molarity \times volume(mL) = 0.1 \times 150 = 15 mmol$

mmoles of $K I$ required = $15 \times 2 = 30 mmol$

Given molarity of $K I = 0.2 M$
Volume of KI required $= \frac{30}{0.2} = 150 mL$

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Similar questions

Q. What minimum volume of 0.2 M

K I (a q)

solution is required to completely precipitate all of the lead in 150 ml

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solution?

2 K I + P b (N O_{3})_{2} (a q) \to 2 K N O_{3} (a q) + P b I_{2} (s)

Given the reaction:-

$P b {(N O_{3})}_{2} (a q) + 2 K I \to P b I_{2} (s) + 2 K N O_{3} (a q)$ What is the mass of $P b I_{2}$ that will precipitate if 10.2 g of $P b {(N O_{3})}_{2}$ is mixed with 5.73 g of $K I$ in a sufficient quantity of $H_{2} O$ ?

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What minimum volume of 0.2 M KI(aq) solution is required to completely precipitate all of the lead in 150 ml 0.1 M Pb(NO3)2 solution? 2KI+Pb(NO3)2(aq)→2KNO3(aq)+PbI2(s)

The correct option is D 150 mLFrom the balanced chemical equation 1 mole of Pb(NO3)2 requires 2 mole of KI. Number of mmoles of Pb(NO3)2 =molarity×volume(mL) =0.1×150=15 mmol mmoles of KI required = 15×2=30 mmol Given molarity of KI=0.2 M Volume of KI required =300.2=150 mL

What minimum volume of 0.2 M $K I (a q)$ solution is required to completely precipitate all of the lead in 150 ml $0.1 M P b (N O_{3})_{2}$ solution? $2 K I + P b (N O_{3})_{2} (a q) \to 2 K N O_{3} (a q) + P b I_{2} (s)$