If the solubility product of PbS is 8 × 10^-28, then the solubility of PbS in pure water at 298 K is $x\times 10^{-16}mol{L}^{-1}.$  The value of x is__________. (Nearest integer)

[Given: $\sqrt[]{2}=1.41]$  

<p>&nbsp;<math><mrow><mi>S</mi><mo>=</mo><mroot><mrow><mi>K</mi><mi>s</mi><mi>p</mi></mrow><mrow></mrow></mroot><mo>=</mo><mroot><mrow><mn>8</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>2</mn><mn>8</mn></mrow></msup></mrow><mrow></mrow></mroot><mo>=</mo><mn>2</mn><mo>.</mo><mn>8</mn><mn>2</mn><mo>×</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>−</mo><mn>1</mn><mn>4</mn></mrow></msup></mrow></math></p><p>Or solubility of PbS = 282 × 10^-16</p>

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