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In order to remove ${Pb}^{2 +}$ from 10 liter $H_{2} O, {Na}_{2} H_{2}$ EDTA (0.4 M, 100 mL ) is required.
${PbCl}_{2} (a q) + {Na}_{2} H_{2} EDTA ⟶ 2 NaCl + {PbH}_{2} EDTA$ Hence millimoles of ${PbCl}_{2}$ present in 1 litre of $H_{2} O$ is :

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answer is 4.

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Detailed Solution

m mole of ${PbCl}_{2}$ in 10 liter of

$\begin{aligned} H_{2} O = 0.4 \times 100 \end{aligned}$

=40.

m mole of ${PbCl}_{2}$ in 1litre of $\begin{aligned} H_{2} = \frac{1 \times 40}{10} \end{aligned}$ $= 4$

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