12. ORGANIC CHEMISTRY SOME BASIC PRINCIPLES AND TECHNIQUES
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GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains \(C = 74.0\% ,\) \(H = 8.65\% \) and \(N = 17.3\% \). Its Empirical formula is

1 \({C_5}{H_8}N\)
2 \({C_{10}}{H_{12}}N\)
3 \({C_5}{H_7}N\)
4 \({C_{10}}{H_{14}}N\)
GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method
2 Diffusion method
3 Vapour pressure method
4 Victor Meyer's method
GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight \(\times\) basicity
2 \(\frac{{{\rm{Equivalent \,weight}}}}{{{\rm{Basicity}}}}\)
3 \(\frac{{{\rm{Basicity}}}}{{{\rm{Equivalent \,weight}}}}\)
4 Equivalent weight \(\times\) valency
GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition
2 Different molecular weight
3 Same viscosity
4 Same vapour density
GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is \({C_2}{H_5}O\) and its molecular weight is \(90\). Molecular formula of the compound is

1 \({C_2}{H_5}O\)
2 \({C_3}{H_6}{O_3}\)
3 \({C_4}{H_{10}}{O_2}\)
4 \({C_5}{H_{14}}O\)
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GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains \(C = 74.0\% ,\) \(H = 8.65\% \) and \(N = 17.3\% \). Its Empirical formula is

1 \({C_5}{H_8}N\)
2 \({C_{10}}{H_{12}}N\)
3 \({C_5}{H_7}N\)
4 \({C_{10}}{H_{14}}N\)
GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method
2 Diffusion method
3 Vapour pressure method
4 Victor Meyer's method
GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight \(\times\) basicity
2 \(\frac{{{\rm{Equivalent \,weight}}}}{{{\rm{Basicity}}}}\)
3 \(\frac{{{\rm{Basicity}}}}{{{\rm{Equivalent \,weight}}}}\)
4 Equivalent weight \(\times\) valency
GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition
2 Different molecular weight
3 Same viscosity
4 Same vapour density
GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is \({C_2}{H_5}O\) and its molecular weight is \(90\). Molecular formula of the compound is

1 \({C_2}{H_5}O\)
2 \({C_3}{H_6}{O_3}\)
3 \({C_4}{H_{10}}{O_2}\)
4 \({C_5}{H_{14}}O\)
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GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains \(C = 74.0\% ,\) \(H = 8.65\% \) and \(N = 17.3\% \). Its Empirical formula is

1 \({C_5}{H_8}N\)
2 \({C_{10}}{H_{12}}N\)
3 \({C_5}{H_7}N\)
4 \({C_{10}}{H_{14}}N\)
GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method
2 Diffusion method
3 Vapour pressure method
4 Victor Meyer's method
GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight \(\times\) basicity
2 \(\frac{{{\rm{Equivalent \,weight}}}}{{{\rm{Basicity}}}}\)
3 \(\frac{{{\rm{Basicity}}}}{{{\rm{Equivalent \,weight}}}}\)
4 Equivalent weight \(\times\) valency
GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition
2 Different molecular weight
3 Same viscosity
4 Same vapour density
GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is \({C_2}{H_5}O\) and its molecular weight is \(90\). Molecular formula of the compound is

1 \({C_2}{H_5}O\)
2 \({C_3}{H_6}{O_3}\)
3 \({C_4}{H_{10}}{O_2}\)
4 \({C_5}{H_{14}}O\)
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GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains \(C = 74.0\% ,\) \(H = 8.65\% \) and \(N = 17.3\% \). Its Empirical formula is

1 \({C_5}{H_8}N\)
2 \({C_{10}}{H_{12}}N\)
3 \({C_5}{H_7}N\)
4 \({C_{10}}{H_{14}}N\)
GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method
2 Diffusion method
3 Vapour pressure method
4 Victor Meyer's method
GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight \(\times\) basicity
2 \(\frac{{{\rm{Equivalent \,weight}}}}{{{\rm{Basicity}}}}\)
3 \(\frac{{{\rm{Basicity}}}}{{{\rm{Equivalent \,weight}}}}\)
4 Equivalent weight \(\times\) valency
GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition
2 Different molecular weight
3 Same viscosity
4 Same vapour density
GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is \({C_2}{H_5}O\) and its molecular weight is \(90\). Molecular formula of the compound is

1 \({C_2}{H_5}O\)
2 \({C_3}{H_6}{O_3}\)
3 \({C_4}{H_{10}}{O_2}\)
4 \({C_5}{H_{14}}O\)
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GENERAL ORGANIC CHEMISTRY

25270 An organic compound contains \(C = 74.0\% ,\) \(H = 8.65\% \) and \(N = 17.3\% \). Its Empirical formula is

1 \({C_5}{H_8}N\)
2 \({C_{10}}{H_{12}}N\)
3 \({C_5}{H_7}N\)
4 \({C_{10}}{H_{14}}N\)
GENERAL ORGANIC CHEMISTRY

25271 An appropriate method for molecular weight determination of chloroform is

1 Regnault's method
2 Diffusion method
3 Vapour pressure method
4 Victor Meyer's method
GENERAL ORGANIC CHEMISTRY

25272 Molecular weight of an organic acid is given by

1 Equivalent weight \(\times\) basicity
2 \(\frac{{{\rm{Equivalent \,weight}}}}{{{\rm{Basicity}}}}\)
3 \(\frac{{{\rm{Basicity}}}}{{{\rm{Equivalent \,weight}}}}\)
4 Equivalent weight \(\times\) valency
GENERAL ORGANIC CHEMISTRY

25273 If two compounds have the same empirical formula but different molecular formulae they must have

1 Different percentage composition
2 Different molecular weight
3 Same viscosity
4 Same vapour density
GENERAL ORGANIC CHEMISTRY

25274 Empirical formula of a compound is \({C_2}{H_5}O\) and its molecular weight is \(90\). Molecular formula of the compound is

1 \({C_2}{H_5}O\)
2 \({C_3}{H_6}{O_3}\)
3 \({C_4}{H_{10}}{O_2}\)
4 \({C_5}{H_{14}}O\)