Question

896 ml of a mixture of CO and CO2 weigh 1.28g at NTP . Calculate the volume of CO2 in the mixture at NTP .
i) 448 ml
ii) 672 ml
iii) 224 ml
iiii) 500 ml

✋actually ☝i m telling u.that its ans. is 224ml
but everyone is finding its ans. as 500 ml which is wrong...

actually i want solution.
Thanks.​

Answer 1

Answer:

224 ml

Explanation:

896 mL of a gaseous mixture weighs 1.28 g. So, assuming ideal gas behaviour and NTP as , we have,

PV=nRT

∴M=wRTPV

∴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≈32 gmol.

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

So, adding up the weights, we have,

28x+44(1−x)≈32

∴44−16x≈32

∴x≈0.75

So, there’s about 25% CO2 in the gaseous mixture by volume or by moles.

So, 25% of 896mL is 224mL.

Answer 2

Answer:

hi Army!!!!!!!!!!!!!!!!