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The Henry's Law constant (k) for carbon monoxide in water at 25°C is 9.71x 10-4 mol/(L-atm). How many grams of CO Will
dissolve in 1.00 L of water if the partial pressure of CO is 2.75 atm?

Sagot :

Erin171idn

In this question, you are given the Henry constant (9.71 × 10-4 mol/l atm), the volume(1 l) and the pressure(2.75 atm). The molecular mass of CO should be 28g/mol.
Then the amount of carbon monoxide in the water in grams should be: 9.71 × 10-4 mol/l atm * 1l * 2.75atm * 28g/mol= 747.67 *10^-4= 0.074767 gramIn this question, you are given the Henry constant (9.71 × 10-4 mol/l atm), the volume(1 l) and the pressure(2.75 atm). The molecular mass of CO should be 28g/mol.
Then the amount of carbon monoxide in the water in grams should be: 9.71 × 10-4 mol/l atm * 1l * 2.75atm * 28g/mol= 747.67 *10^-4= 0.074767 grams


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