Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg. ? sir plz solution bata dijiye
Given: KH = 4.27×10^5 mmHg P = 760 mmHg To find: solubility of methane in benzene (x) = ? According to Henry's law, P = KH × x Therefore, x = P/KH X = 760/4.27×10^5 X = 0.001779 Or X = 1.78×10^-3 (solubility of methane in benzene)
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Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg. ?
sir plz solution bata dijiye
ans is 1.4.10^3 g of methane
Given:
KH = 4.27×10^5 mmHg
P = 760 mmHg
To find: solubility of methane in benzene (x) = ?
According to Henry's law,
P = KH × x
Therefore, x = P/KH
X = 760/4.27×10^5
X = 0.001779
Or
X = 1.78×10^-3 (solubility of methane in benzene)
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