(3) 10 of 10 ( 0.72 g/(100 mL) 00 mL) Correct The Henry\'s law equation, C/P Ca/

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(3) 10 of 10 ( 0.72 g/(100 mL) 00 mL) Correct The Henry's law equation, C/P Ca/P, allows you to calculate the solubility when you are given values for throe of the four known vanables. Ci , P,"and P,inthe equation Ci/p=G/P:. Ci is the solubilty of CO2 in water. C2 is the solubity of CO2 in a soft drink, Pt is the pressure of CO2 in water, and P is the pressure of the CO2 in a soft drink. The known quantities are Ci. Pi, and Py. First,rostate the Henry's law equation to isolate the unknown variable, C Yuare g en G-0.15 g/(100 mL),A=760 mmHg, and F, 4.8 atrn. Since P, and n are in two dtrent units, use the equality 760 mmHg = 1 atm to convert them in the same unit of atm: 760 urig--= 1 atm G-1(015 g) (100ml )1 48 . (0.72 g)/(100 mL) An atmospheric concentration of 380 ppm CO2 corresponds to dissolved in the solution in Part A remains in solution after the soft drink a partial pressure of 0.00038 atm. What percentage of the CO2 originally MacBook Pro

Explanation / Answer

H= Henry's law constant for CO2= 29 L.atm/mole

H= P/Caq, P = partial pressure of CO2 and Caq= concentration of CO2 in the aqueous phase

Caq= P/H= 0.00038 atm/29 mole/L=1.31*10-5 moles/L

1 L = 1000ml

1000 ml contains 1.31*10-5 moles of CO2= 1.31*10-5*44 gm of CO2 ( mass= moles* molar mass and molar mass of CO2= 44 g/mole )=0.00058 gm

100ml contains 0.00058*100/1000 ml =0.000058 gm in 100 ml

originally there is 0.72 gm/100ml

% of CO2 present in solution after equilibrium=100*( 0.000058/100ml/0.72/100)=0.081%

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