henry's law constant for solubility of methane in benzene is 4.27×10 mm hg at constant temperature. calculate solubility of methane at 760 mm of hg pressure at same temperature.
Answers
Answer:
In chemistry, Henry's law is a gas law that states that the amount of dissolved gas is proportional to its partial pressure in the gas phase. The proportionality factor is called the Henry's law constant.
In chemistry, Henry's law is a gas law that states that the amount of dissolved gas is proportional to its partial pressure in the gas phase. The proportionality factor is called the Henry's law constant.An everyday example of Henry’s law is given by carbonated soft drinks. Before the bottle or can is opened, the gas above the drink is almost pure carbon dioxide at a pressure slightly higher than atmospheric pressure. The drink itself contains dissolved carbon dioxide. When the bottle or can is opened, some of this gas escapes, giving the characteristic hiss (or pop in the case of a sparkling wine bottle). Because the pressure above the liquid is now lower, some of the dissolved carbon dioxide comes out of solution as bubbles. If a glass of the drink is left in the open, the concentration of carbon dioxide in solution will come into equilibrium with the carbon dioxide in the air, and the drink will go flat.
Explanation:
Given that Henry’s law constant
Given that Henry’s law constantKH=4.27×10^5mmHg
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHg
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction then
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX760=4.27×105×X
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX760=4.27×105×X Divide by 4.27 × 105 ,we get
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX760=4.27×105×X Divide by 4.27 × 105 ,we getMolar fraction, X
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX760=4.27×105×X Divide by 4.27 × 105 ,we getMolar fraction, X =760 ÷ 4.27×10^5
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX760=4.27×105×X Divide by 4.27 × 105 ,we getMolar fraction, X =760 ÷ 4.27×10^5 =178×10^-5
Given that Henry’s law constantKH=4.27×10^5mmHgp=760mmHgIf X is the molar fraction thenAccording to Henry’s law,p=KHX760=4.27×105×X Divide by 4.27 × 105 ,we getMolar fraction, X =760 ÷ 4.27×10^5 =178×10^-5So that, the molar fraction of methane in benzene =1.78×10^-3.
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