A) 4.6 g
B) 1.6 g
C) 2.3 g
D) 23 g
Correct Answer: C
Solution :
[c] Equimolecular proportion means both gases occupied equal volume \[=\frac{2.24}{2}=1.12L\] |
For \[C{{H}_{4}}\]: |
22.4L \[C{{H}_{4}}\]has mass \[=16gm\] |
1.12L \[C{{H}_{4}}\]has mass \[=\frac{16}{22.4}\times 1.12=0.8gm\]. |
For \[{{C}_{2}}{{H}_{6}}\] |
22.4L \[{{C}_{2}}{{H}_{6}}\]has mass = 30gm |
1.12L\[{{C}_{2}}{{H}_{6}}\]has mass \[=\frac{30}{22.4}\times 1.12\]\[=\frac{3.0}{2}gm=1.5gm\] |
Total mass \[=1.5gm+0.8gm=2.3gm\]. |
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