306635
Given the following percentage composition: Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?
306636
0.145 g of hydrocarbon is heated with dry copper (II) oxide and \({\rm{224}}{\mkern 1mu} {\mkern 1mu} {\rm{c}}{{\rm{m}}^{\rm{3}}}\) of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) was collected at STP. The empirical formula of the hydrocarbon is
306637
A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
3 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
4 \({{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)
Explanation:
Given, % of H = 12.5% \(\therefore \% \,\,of\,\,N = 100 - 12.5 = 87.5\% \)2\( \times \) vapour density = Mol.wt \(Mol.wt. = 16 \times 2 = 32\) Molecular formula \( = n \times empirical\,\,formula\,\,mass\) \(n = \frac{{32}}{{16}} = 2\) \(\therefore \) Molecular formula of the compound will be \( = {(N{H_2})_2} = {N_2}{H_4}\)
JEE - 2014
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306638
Statement A : The empirical mass of ethyne is half of its molecular mass. Statement B : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
306635
Given the following percentage composition: Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?
306636
0.145 g of hydrocarbon is heated with dry copper (II) oxide and \({\rm{224}}{\mkern 1mu} {\mkern 1mu} {\rm{c}}{{\rm{m}}^{\rm{3}}}\) of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) was collected at STP. The empirical formula of the hydrocarbon is
306637
A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
3 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
4 \({{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)
Explanation:
Given, % of H = 12.5% \(\therefore \% \,\,of\,\,N = 100 - 12.5 = 87.5\% \)2\( \times \) vapour density = Mol.wt \(Mol.wt. = 16 \times 2 = 32\) Molecular formula \( = n \times empirical\,\,formula\,\,mass\) \(n = \frac{{32}}{{16}} = 2\) \(\therefore \) Molecular formula of the compound will be \( = {(N{H_2})_2} = {N_2}{H_4}\)
JEE - 2014
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306638
Statement A : The empirical mass of ethyne is half of its molecular mass. Statement B : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306635
Given the following percentage composition: Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?
306636
0.145 g of hydrocarbon is heated with dry copper (II) oxide and \({\rm{224}}{\mkern 1mu} {\mkern 1mu} {\rm{c}}{{\rm{m}}^{\rm{3}}}\) of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) was collected at STP. The empirical formula of the hydrocarbon is
306637
A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
3 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
4 \({{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)
Explanation:
Given, % of H = 12.5% \(\therefore \% \,\,of\,\,N = 100 - 12.5 = 87.5\% \)2\( \times \) vapour density = Mol.wt \(Mol.wt. = 16 \times 2 = 32\) Molecular formula \( = n \times empirical\,\,formula\,\,mass\) \(n = \frac{{32}}{{16}} = 2\) \(\therefore \) Molecular formula of the compound will be \( = {(N{H_2})_2} = {N_2}{H_4}\)
JEE - 2014
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306638
Statement A : The empirical mass of ethyne is half of its molecular mass. Statement B : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
306635
Given the following percentage composition: Na = 29.11%, S = 40.51% and O = 30.38%. If molar mass of the compound is 158, what is its molecular formula?
306636
0.145 g of hydrocarbon is heated with dry copper (II) oxide and \({\rm{224}}{\mkern 1mu} {\mkern 1mu} {\rm{c}}{{\rm{m}}^{\rm{3}}}\) of \({\rm{C}}{{\rm{O}}_{\rm{2}}}\) was collected at STP. The empirical formula of the hydrocarbon is
306637
A gaseous compound of nitrogen and hydrogen contains 12.5 % by mass of hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is:
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
3 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
4 \({{\rm{N}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)
Explanation:
Given, % of H = 12.5% \(\therefore \% \,\,of\,\,N = 100 - 12.5 = 87.5\% \)2\( \times \) vapour density = Mol.wt \(Mol.wt. = 16 \times 2 = 32\) Molecular formula \( = n \times empirical\,\,formula\,\,mass\) \(n = \frac{{32}}{{16}} = 2\) \(\therefore \) Molecular formula of the compound will be \( = {(N{H_2})_2} = {N_2}{H_4}\)
JEE - 2014
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306638
Statement A : The empirical mass of ethyne is half of its molecular mass. Statement B : The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
