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The minimum volume of water required to dissolve 0.1 g lead(II) chloride to get a saturated solution( (K_{sp} of PbCI_{2} = 3.2\times 10^{-8} ; automic mass of Pb =207 u) is :

  • Option 1)

    0.36 L

  • Option 2)

    17.98 L

  • Option 3)

    0.18 L

  • Option 4)

    1.798 L

 

Answers (1)

best_answer

As we learned

(K_{sp})PbCI_{2} = 32\times 10^{9}

PbCI_{2}\rightleftharpoons Pb^{2}+2CI^{-}

                        3            2s

K_{sp}=[Pb^{+2}][a^{-}]^{2}

K_{sp}=[4s^{3}]=32\times 10^{-9}

s^{3}= 8\times 10^{-9}

3=2\times 10^{-3}

\frac{weight}{mol cot}\times \frac{1}{V_{1}}=2\times 10^{-3}

\frac{0.1}{278}\times \frac{1}{V_{2}}=2\times 10^{-3}

V_{2}=\frac{0.1\times 1000}{278\times 2}=0.18L


Option 1)

0.36 L

Option 2)

17.98 L

Option 3)

0.18 L

Option 4)

1.798 L

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Plabita

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