2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the same pressure. What is the molar mass of the gas?
From the gas equation,
PV = (w/M) RT
Substituting the given data in the gas equation, we get
PV = (2.9 / M) x R x 368
&
PV = (0.184 / 2) x R x 290
From these two equation, we can write
(2.9 / M) x R x 368 = (0.184 / 2) x R x 290
By, striking throug R from both side, we get
(2.9 / M) x 368 = (0.184 / 2) x 290
Or
(2.9 / M) = (0.092 X 290) / 368
Or
M = 2.9 x 368 / 0.092 x 290
= 40 g/mol
Hence, the molar mass of the gas is 40 g mol–1.
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 18: 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C, at the s....
Comments
plz specify
why are there so many 'or's , its little confusing
39.67
38.94g/mol
What is w÷M i dont hear this