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QAIPM-2011-Ch-1866: NEET 2011 Chemistry - Class 11 AIPM-2011-Ch-1866 Solution: A gaseous mixture was prepared by taking equal mole of CO and…
A gaseous mixture was prepared by taking equal mole of CO and . If the total pressure of the mixture was found 1 atmosphere the partial pressure of the nitrogen () in the mixture is 0.5 atm 0.8 atm 0.9 atm 1 atm
Multiple Choice Options
- A 15.45
- B 16.45
- C 17.45
- D 14.45
Step-by-Step Verified Solution
Given V 1
= 55 mL, V 2
= ?
P 1
= 715 � 15 = 700 mm, P 2
= 760 mm
T 1
= 300 K, T 2
= 273 K
General gas equation,
P 1 V 1 T 1 = P 2 V 2 T 2
_ 700 _ 55 300 = 760 _ V 2 273
_ V 2 = 46.098 mL
Now, 22400 mL of nitrogen = 1 mole
_ 46.098 mL of nitrogen = 1 _ 46.098 22400 mol
Weight of nitrogen = 1 _ 46.098 22400 _ 28 = 0.057 g
Percent composition of nitrogen in 0.35 g of
compound
= 0.0576 0.35 _ 100 = 16.45 %