InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Arrange the following in the order of increasing mass (at. Mass of O=16, Cu=63, N=14) (I) one atom of oxygen (II) one atom of nitrogen (III) `1xx10^(-10)` mole of oxygen (IV) `1xx10^(-10)` mole of copperA. II lt I lt III lt IVB. I lt II lt III lt IVC. III lt II lt IV lt ID. IV lt II lt III lt I |
|
Answer» Correct Answer - A (i) 1 atom of oxygen `=(16)/(6.02xx10^(23)g)` `=2.65xx10^(-23)g` (ii) 1 atom of nitrogen `=(14)/(6.02xx10^(23))g` `=2.326xx10^(-23)g` (iii) `1xx10^(-10)` mol of oxygen `=32xx10^(-10)` `=3.2xx10^(-9)`g (iv) `1xx10^(-10)` mol of `CO=63.5xx10^(-10)g` `=6.35xx10^(-9)`g Thus, the increaseing order of mass is II lt I lt III lt IV |
|