InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
10 mL of H2 combine with 5 mL of O2 to form water. When 200 mL of H2 at STP is passed over heated CuO, the CuO loses 0.144 g of its weight. Does the above data correspond to the law of constant composition? |
|
Answer» In the second reaction, 0.144 g weight is lost from CuO which is due to reduction of CuO into Cu. We can say that 0.144 g oxygen is combined with 200 mL H2. \(\because\) 32 g oxygen occupies 22400 mL volume at STP \(\therefore\) 0.144 g oxygen will occupy = \(\frac{2240}{32}\) x 0.144 = 100.8 mL of O2 \(\therefore\) Ratio of H2 and O2 in water = 200 : 100.8 = 2 : 1 This ratio is equivalent to first reaction (i.e. 10 : 5 or 2 : 1). Hence, the above data corresponds to the law of constant composition. |
|