896 mL of a mixture of CO2 and CO, weigh 1.28 g
at NTP. Calculate the volume of CO2, in the mixture at NTP
at NTP.
(1) 448 ml
(2) 672 ml
(3) 224 ml
(4) 500 ml
Answers
Answer:
option C - 224 ml
Explanation:
Let the weight of CO in the mixture be x g.
Total weight of mixture =1.28g
Weight of CO
2
=1.28−x
Moleculr weight of CO=28g
Molecular weight of CO
2
=44g
Moles of CO
2
in the mixture =
44
1.28−x
Moles of CO in the mixture =
28
x
At NTP, the volume occupied by 1 mole of a gas =22400mL
∴ 896 mL occupied by
22400
1
×896=0.04 moles
∴ Total moles in the mixture = 0.04 moles
⇒ Moles of CO
2
+ Moles of CO=0.04
⇒
44
1.28−x
+
28
x
=0.04
⇒35.84−28x+44x=49.28
⇒16x=13.44
⇒x=0.84
∴ Molecular weight of CO
2
=1.28−0.84=0.44g
∴ Moles of CO
2
in the mixture =
44
0.44
=0.01 moles
∵ Volume occupied by 1 mole of CO
2
=22400mL
⇒ Volume occupied by 0.01 mole of CO
2
=22400×0.01=224mL
Hence, at NTP, 224 mL of CO
2
is present in the mixture.