In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0⁰C and atmospheric pressure. Calculate (i) the mass fraction and (ii) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 and water.Basis 1 m3 of water = 1000 kgVolume of carbon dioxide added = 3 m3From the gas equation, pV = nRT1 x 3 = n x 0.08206 x 273 n = 0.134 mole.Molecular weight of carbon dioxide = 44And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg

In the carbonation of a soft drink, the total quantity of carbon diox

1.	In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0⁰C and atmospheric pressure. Calculate (i) the mass fraction and (ii) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 and water.Basis 1 m3 of water = 1000 kgVolume of carbon dioxide added = 3 m3From the gas equation, pV = nRT1 x 3 = n x 0.08206 x 273 n = 0.134 mole.Molecular weight of carbon dioxide = 44And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg
A.	6.9 × 10-3, 3.41 × 10-3
B.	8.9 × 10-3, 2.41 × 10-3
C.	5.9 × 10-3, 2.41 × 10-3
D.	9.9 × 10-3, 3.41 × 10-3
Answer» D. 9.9 × 10-3, 3.41 × 10-3

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In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0⁰C and atmospheric pressure. Calculate (i) the mass fraction and (ii) the mole fraction of the CO2 in the drink, ignoring all components other than CO2 and water.Basis 1 m3 of water = 1000 kgVolume of carbon dioxide added = 3 m3From the gas equation, pV = nRT1 x 3 = n x 0.08206 x 273 n = 0.134 mole.Molecular weight of carbon dioxide = 44And so weight of carbon dioxide added = 0.134 x 44 = 5.9 kg
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