A) \[2.5\times {{10}^{-3}}\]
B) \[2.0\times {{10}^{-3}}\]
C) \[1.25\times {{10}^{-3}}\]
D) \[5.0\times {{10}^{-3}}\]
Correct Answer: A
Solution :
As \[4\times {{10}^{-3}}kg\]of gas is dissolved in 1 L of water. According to Henry s law \[m={{K}_{H}}p\] Where, \[m=4\times {{10}^{-3}}kg\] and \[p=100atm\] \[\therefore \] \[{{K}_{H}}=\frac{m}{p}-=\frac{4\times {{10}^{-3}}}{100}\] \[=4\times {{10}^{-4}}kg{{L}^{-1}}at{{m}^{-1}}\] Now. if the pressure is increased to 250 atm the mass of gas\[(m)={{K}_{H}}p\] \[=4\times {{10}^{-5}}\times 250\] \[={{10}^{-2}}kg\,{{L}^{-1}}\] and the mass of gas present in 250 mL \[=\frac{250}{100}\times {{10}^{-2}}\] \[=2.5\times {{10}^{-3}}kg\]
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