Question

896 ml of a mixture of CO and CO2 weigh 1.28 g at NTP . calculate the volume of co2 in the mixture

Answer 1

PV=nRTPV=nRT

∴M=wRTPV∴M=wRTPV

∴M=1.28 g×0.0821 L⋅atmmol⋅K×273.15 K1 atm×896 mL×11000×LmL∴M=1.28 g×0.0821 L⋅atmmol⋅K×273.15 K1 atm×896 mL×11000×LmL

∴M=1.28×0.0821×273.15×10001×896 gmol∴M=1.28×0.0821×273.15×10001×896 gmol

∴M≈32 gmol∴M≈32 gmol.

So, 1 mol1 mol gaseous mixture weighs 32 g32 g. Let 1 mol1 mol of the gaseous mixture contain x mol COx mol CO. So, it contains (1−x) mol CO2(1−x) mol CO2.

So, adding up the weights, we have,

28x+44(1−x)≈3228x+44(1−x)≈32

∴44−16x≈32∴44−16x≈32

∴x≈0.75∴x≈0.75

So, there’s about 25% CO2CO2 in the gaseous mixture by volume or by moles.

So, 25% of 896mL896mL is 224mL224mL.

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