NEET Test Series from KOTA - 10 Papers In MS WORD
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GENERAL ORGANIC CHEMISTRY
25266
On complete combustion \(1.4 \,gm\) hydrocarbon gave \(1.8\, gm\) water. Empirical formula of the hydrocarbon is
1 \(CH\)
2 \(C{H_2}\)
3 \(C{H_3}\)
4 \(C{H_4}\)
Explanation:
(b) \(1.8\,gm\) water obtained from \(1.4\,gm\) hydrocarbon \(\therefore \) \(18\,gm\) water obtained from -\(\frac{{1.4}}{{1.8}} \times 18 = 14\) \(gm\). Empirical formula Mass = \(14\) \(\therefore \) Empirical formula = \(C{H_2}\).
GENERAL ORGANIC CHEMISTRY
25267
In the estimation of sulphur organic compound on treating with conc. \(HN{O_3}\) is converted to
1 \(S{O_2}\)
2 \({H_2}S\)
3 \({H_2}S{O_4}\)
4 \(S{O_3}\)
Explanation:
(c)In carius method sulphur of organic compound is converted in to \({H_2}S{O_4}\) \(S+{{H}_{2}}O+3O\,\underset{HN{{O}_{3}}}{\mathop{\xrightarrow{\Delta }}}\,{{H}_{2}}S{{O}_{4}}\)
GENERAL ORGANIC CHEMISTRY
25268
In Carius method \(0.099 \,g\) organic compound gave \(0.287\, g\) \(AgCl\). The percentage of chlorine in the compound will be
25269
\(0.24 \,g\) of an organic compound gave \(0.22\, g\) \(C{O_2}\) on complete combustion. If it contains \(1.66\%\) hydrogen, then the percentage of \(C\) and \(O\) will be
1 \(12.5\) and \(36.6\)
2 \(25\) and \(75\)
3 \(25\) and \(36.6\)
4 \(25\) and \(80\)
Explanation:
(b) \(\%\) of \(C\) = \(\frac{{12}}{{44}} \times \frac{{{\text{Mass of }}C{O_2}}}{{{\text{Mass of substance}}}} \times 100\) = \(\frac{{12 \times 0.22}}{{44 \times 0.24}} \times 100 = 25; \,C = 25, \,H = 1.66\) Total = \(26.6 = 100 -26.6 = 73.4.\)
25266
On complete combustion \(1.4 \,gm\) hydrocarbon gave \(1.8\, gm\) water. Empirical formula of the hydrocarbon is
1 \(CH\)
2 \(C{H_2}\)
3 \(C{H_3}\)
4 \(C{H_4}\)
Explanation:
(b) \(1.8\,gm\) water obtained from \(1.4\,gm\) hydrocarbon \(\therefore \) \(18\,gm\) water obtained from -\(\frac{{1.4}}{{1.8}} \times 18 = 14\) \(gm\). Empirical formula Mass = \(14\) \(\therefore \) Empirical formula = \(C{H_2}\).
GENERAL ORGANIC CHEMISTRY
25267
In the estimation of sulphur organic compound on treating with conc. \(HN{O_3}\) is converted to
1 \(S{O_2}\)
2 \({H_2}S\)
3 \({H_2}S{O_4}\)
4 \(S{O_3}\)
Explanation:
(c)In carius method sulphur of organic compound is converted in to \({H_2}S{O_4}\) \(S+{{H}_{2}}O+3O\,\underset{HN{{O}_{3}}}{\mathop{\xrightarrow{\Delta }}}\,{{H}_{2}}S{{O}_{4}}\)
GENERAL ORGANIC CHEMISTRY
25268
In Carius method \(0.099 \,g\) organic compound gave \(0.287\, g\) \(AgCl\). The percentage of chlorine in the compound will be
25269
\(0.24 \,g\) of an organic compound gave \(0.22\, g\) \(C{O_2}\) on complete combustion. If it contains \(1.66\%\) hydrogen, then the percentage of \(C\) and \(O\) will be
1 \(12.5\) and \(36.6\)
2 \(25\) and \(75\)
3 \(25\) and \(36.6\)
4 \(25\) and \(80\)
Explanation:
(b) \(\%\) of \(C\) = \(\frac{{12}}{{44}} \times \frac{{{\text{Mass of }}C{O_2}}}{{{\text{Mass of substance}}}} \times 100\) = \(\frac{{12 \times 0.22}}{{44 \times 0.24}} \times 100 = 25; \,C = 25, \,H = 1.66\) Total = \(26.6 = 100 -26.6 = 73.4.\)
25266
On complete combustion \(1.4 \,gm\) hydrocarbon gave \(1.8\, gm\) water. Empirical formula of the hydrocarbon is
1 \(CH\)
2 \(C{H_2}\)
3 \(C{H_3}\)
4 \(C{H_4}\)
Explanation:
(b) \(1.8\,gm\) water obtained from \(1.4\,gm\) hydrocarbon \(\therefore \) \(18\,gm\) water obtained from -\(\frac{{1.4}}{{1.8}} \times 18 = 14\) \(gm\). Empirical formula Mass = \(14\) \(\therefore \) Empirical formula = \(C{H_2}\).
GENERAL ORGANIC CHEMISTRY
25267
In the estimation of sulphur organic compound on treating with conc. \(HN{O_3}\) is converted to
1 \(S{O_2}\)
2 \({H_2}S\)
3 \({H_2}S{O_4}\)
4 \(S{O_3}\)
Explanation:
(c)In carius method sulphur of organic compound is converted in to \({H_2}S{O_4}\) \(S+{{H}_{2}}O+3O\,\underset{HN{{O}_{3}}}{\mathop{\xrightarrow{\Delta }}}\,{{H}_{2}}S{{O}_{4}}\)
GENERAL ORGANIC CHEMISTRY
25268
In Carius method \(0.099 \,g\) organic compound gave \(0.287\, g\) \(AgCl\). The percentage of chlorine in the compound will be
25269
\(0.24 \,g\) of an organic compound gave \(0.22\, g\) \(C{O_2}\) on complete combustion. If it contains \(1.66\%\) hydrogen, then the percentage of \(C\) and \(O\) will be
1 \(12.5\) and \(36.6\)
2 \(25\) and \(75\)
3 \(25\) and \(36.6\)
4 \(25\) and \(80\)
Explanation:
(b) \(\%\) of \(C\) = \(\frac{{12}}{{44}} \times \frac{{{\text{Mass of }}C{O_2}}}{{{\text{Mass of substance}}}} \times 100\) = \(\frac{{12 \times 0.22}}{{44 \times 0.24}} \times 100 = 25; \,C = 25, \,H = 1.66\) Total = \(26.6 = 100 -26.6 = 73.4.\)
25266
On complete combustion \(1.4 \,gm\) hydrocarbon gave \(1.8\, gm\) water. Empirical formula of the hydrocarbon is
1 \(CH\)
2 \(C{H_2}\)
3 \(C{H_3}\)
4 \(C{H_4}\)
Explanation:
(b) \(1.8\,gm\) water obtained from \(1.4\,gm\) hydrocarbon \(\therefore \) \(18\,gm\) water obtained from -\(\frac{{1.4}}{{1.8}} \times 18 = 14\) \(gm\). Empirical formula Mass = \(14\) \(\therefore \) Empirical formula = \(C{H_2}\).
GENERAL ORGANIC CHEMISTRY
25267
In the estimation of sulphur organic compound on treating with conc. \(HN{O_3}\) is converted to
1 \(S{O_2}\)
2 \({H_2}S\)
3 \({H_2}S{O_4}\)
4 \(S{O_3}\)
Explanation:
(c)In carius method sulphur of organic compound is converted in to \({H_2}S{O_4}\) \(S+{{H}_{2}}O+3O\,\underset{HN{{O}_{3}}}{\mathop{\xrightarrow{\Delta }}}\,{{H}_{2}}S{{O}_{4}}\)
GENERAL ORGANIC CHEMISTRY
25268
In Carius method \(0.099 \,g\) organic compound gave \(0.287\, g\) \(AgCl\). The percentage of chlorine in the compound will be
25269
\(0.24 \,g\) of an organic compound gave \(0.22\, g\) \(C{O_2}\) on complete combustion. If it contains \(1.66\%\) hydrogen, then the percentage of \(C\) and \(O\) will be
1 \(12.5\) and \(36.6\)
2 \(25\) and \(75\)
3 \(25\) and \(36.6\)
4 \(25\) and \(80\)
Explanation:
(b) \(\%\) of \(C\) = \(\frac{{12}}{{44}} \times \frac{{{\text{Mass of }}C{O_2}}}{{{\text{Mass of substance}}}} \times 100\) = \(\frac{{12 \times 0.22}}{{44 \times 0.24}} \times 100 = 25; \,C = 25, \,H = 1.66\) Total = \(26.6 = 100 -26.6 = 73.4.\)
