Question

896 ml of mixture of co and co2 weights 1.28g . 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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Answer 2

Answer:

224 ml

Explanation: Let's assume 'x' ml CO2

and hence '896-x' ml CO.

Also assume 'y' g CO2

and hence '1.28-y' g CO.

by equating moles of CO2 i.e.

x/22400= y/44

and by equating moles of CO i.e.

896-x/22400 = 1.28-y/28

now by solving the two equations two variables we get y=11/25 and therefore x= 224 ml.