| Compound (molecular wt.) | Weight of compound I (in g) taken |
| (i) \[C{{O}_{2}}(44)\] | 4.4 |
| (ii) \[N{{O}_{2}}(46)\] | 2.3 |
| (iii) \[{{H}_{2}}{{O}_{2}}(34)\] | 6.8 |
| (iv) \[S{{O}_{2}}(64)\] | 1.6 |
A) II and IV
B) I and III
C) I and II
D) III and IV
Correct Answer: A
Solution :
In \[\text{C}{{\text{O}}_{2}}\] Molecular wt. \[=12+2\times 16\] \[=12+32=44\] \[\because \] \[44\,g\,C{{O}_{2}}\]contains \[{{O}_{2}}=32\,g\] \[\therefore \] \[4.4\,g\,C{{O}_{2}}\]contains \[{{O}_{2}}=\frac{32}{44}\times 4.4=3.2\,g\] In \[N{{O}_{2}},\] Molecular wt. \[=14+32=46\] \[\because \] \[46\,g\]of \[\text{N}{{\text{O}}_{\text{2}}}\]contains \[{{\text{O}}_{2}}=32\,g\] \[\therefore \] \[2.3\,g\]of \[\text{N}{{\text{O}}_{\text{2}}}\]contains \[{{O}_{2}}=\frac{32\times 2.3}{46}=1.6\,g\] In \[{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\] Molecular wt. \[\text{= 2}\,\,\text{+ 32 =}\] \[\because \]\[\text{34}\,\text{g}\,{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}\]contains \[{{\text{O}}_{2}}=32\,\,g\] \[\therefore \]\[6.8\,g\,{{H}_{2}}{{O}_{2}}\]contains \[{{O}_{2}}=\frac{32.\times 6.8}{34}=4.6\,g\] In \[\text{S}{{\text{O}}_{\text{2}}}\] Molecular wt. \[\text{= 32}\,\text{+}\,\text{32}\,\text{=}\,\text{64}\] \[\because \] 64 g of \[\text{S}{{\text{O}}_{\text{2}}}\]contains \[{{\text{O}}_{2}}=32\,g\] \[\therefore \]\[1.6\,g\,S{{O}_{2}}\]contains \[{{O}_{2}}=\frac{32\times 1.6}{64}=0.8\,g\] Hence, molecule II and IV (\[\text{N}{{\text{O}}_{\text{2}}}\]and\[\text{S}{{\text{O}}_{\text{2}}}\]) have least weight of oxygen.
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