8g CH4 occupies 11.2 dm3 at s.t.p. What volume would 22 g of CH3CH2CH3 occupy under the same condition?
(C = 12, H = 1)
- 3.7 dm3
-
11.2 dm3
- 22.4 dm3
- 33.6 dm3
The correct answer is: B
Explanation
1 mole of any gas has a volume of 22.4dm3 at STP, SO 1mole of propane = 44g = 22.4dm3
22g = ?
( 22 x 22.4)/ 44 = 11.2dm3