306643
What is the molecular mass of a substance which contains 9 carbon atoms,13 hydrogen atoms and \({\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) g of other component?
1 135
2 108
3 13
4 144
Explanation:
The molecule has C, H and other component. Mass of 9 C atoms = 12 \( \times \) 9 = 108 amu Mass of 13 H atoms = 13 \( \times \) 1 = 13 amu Mass of other component \({\rm{ = }}\frac{{{\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}}}{{{\rm{1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}}}{\rm{ = 14}}{\rm{.04 = 135}}{\rm{.04}}\,\,{\rm{amu}}\) \(\therefore \) Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu \(\therefore \) Mol. mass of substance = 135.04
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306644
A compound contains Nitrogen and hydrogen. Its elemental analysis gave 87.5% N. The correct formula of the compound based on the given data is
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
3 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
4 \({\rm{NH}}\)
Explanation:
Hence, the formula of the compound is \({\rm{N}}{{\rm{H}}_{\rm{2}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306645
\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is
306646
A compound having the formula \({\rm{B}}{{\rm{r}}_{\rm{3}}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{3}}}{\left( {{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}} \right)_{\rm{n}}}\) contains \(10.46\,\% \,\,{\rm{Br}}\) by mass; the value of n is
1 54
2 65
3 45
4 35
Explanation:
10.46 g Br is present in 100 g compound \(3 \times 80 \mathrm{~g}\) ' Br ' is present in? \( = \frac{{3 \times 80 \times 100}}{{10.46}} = 2294.45\,\,{\text{gram}}\) \(\therefore\) Mass of \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)_{\text {n }}\) \(=2294.45-[(3 \times 80)+(6 \times 12)+(3 \times 1)]\) \(=1979.45\) \(\therefore \mathrm{n}=\dfrac{1979.45}{44}=45\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306647
A sample of pure compound contains \(\mathrm{1.15 \mathrm{~g}}\) of sodium, \(\mathrm{3.01 \times 10^{22}}\) atoms of carbon and \(\mathrm{0.1 \mathrm{~mol}}\) of oxygen atoms. Its empirical formula is
1 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{3}}\)
2 \(\mathrm{\mathrm{NaCO}_{2}}\)
3 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}}\)
4 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{2}}\)
Explanation:
Moles of \(\mathrm{\mathrm{Na}=\dfrac{1.15 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{C}=\dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{O}=0.1}\) Moles ratio, \(\mathrm{\mathrm{Na}: \mathrm{C}: \mathrm{O}=1: 1: 2}\) \(\mathrm{\therefore}\) Empirical formula \(\mathrm{-\mathrm{NaCO}_{2}}\)
306643
What is the molecular mass of a substance which contains 9 carbon atoms,13 hydrogen atoms and \({\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) g of other component?
1 135
2 108
3 13
4 144
Explanation:
The molecule has C, H and other component. Mass of 9 C atoms = 12 \( \times \) 9 = 108 amu Mass of 13 H atoms = 13 \( \times \) 1 = 13 amu Mass of other component \({\rm{ = }}\frac{{{\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}}}{{{\rm{1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}}}{\rm{ = 14}}{\rm{.04 = 135}}{\rm{.04}}\,\,{\rm{amu}}\) \(\therefore \) Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu \(\therefore \) Mol. mass of substance = 135.04
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306644
A compound contains Nitrogen and hydrogen. Its elemental analysis gave 87.5% N. The correct formula of the compound based on the given data is
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
3 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
4 \({\rm{NH}}\)
Explanation:
Hence, the formula of the compound is \({\rm{N}}{{\rm{H}}_{\rm{2}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306645
\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is
306646
A compound having the formula \({\rm{B}}{{\rm{r}}_{\rm{3}}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{3}}}{\left( {{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}} \right)_{\rm{n}}}\) contains \(10.46\,\% \,\,{\rm{Br}}\) by mass; the value of n is
1 54
2 65
3 45
4 35
Explanation:
10.46 g Br is present in 100 g compound \(3 \times 80 \mathrm{~g}\) ' Br ' is present in? \( = \frac{{3 \times 80 \times 100}}{{10.46}} = 2294.45\,\,{\text{gram}}\) \(\therefore\) Mass of \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)_{\text {n }}\) \(=2294.45-[(3 \times 80)+(6 \times 12)+(3 \times 1)]\) \(=1979.45\) \(\therefore \mathrm{n}=\dfrac{1979.45}{44}=45\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306647
A sample of pure compound contains \(\mathrm{1.15 \mathrm{~g}}\) of sodium, \(\mathrm{3.01 \times 10^{22}}\) atoms of carbon and \(\mathrm{0.1 \mathrm{~mol}}\) of oxygen atoms. Its empirical formula is
1 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{3}}\)
2 \(\mathrm{\mathrm{NaCO}_{2}}\)
3 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}}\)
4 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{2}}\)
Explanation:
Moles of \(\mathrm{\mathrm{Na}=\dfrac{1.15 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{C}=\dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{O}=0.1}\) Moles ratio, \(\mathrm{\mathrm{Na}: \mathrm{C}: \mathrm{O}=1: 1: 2}\) \(\mathrm{\therefore}\) Empirical formula \(\mathrm{-\mathrm{NaCO}_{2}}\)
306643
What is the molecular mass of a substance which contains 9 carbon atoms,13 hydrogen atoms and \({\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) g of other component?
1 135
2 108
3 13
4 144
Explanation:
The molecule has C, H and other component. Mass of 9 C atoms = 12 \( \times \) 9 = 108 amu Mass of 13 H atoms = 13 \( \times \) 1 = 13 amu Mass of other component \({\rm{ = }}\frac{{{\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}}}{{{\rm{1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}}}{\rm{ = 14}}{\rm{.04 = 135}}{\rm{.04}}\,\,{\rm{amu}}\) \(\therefore \) Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu \(\therefore \) Mol. mass of substance = 135.04
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306644
A compound contains Nitrogen and hydrogen. Its elemental analysis gave 87.5% N. The correct formula of the compound based on the given data is
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
3 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
4 \({\rm{NH}}\)
Explanation:
Hence, the formula of the compound is \({\rm{N}}{{\rm{H}}_{\rm{2}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306645
\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is
306646
A compound having the formula \({\rm{B}}{{\rm{r}}_{\rm{3}}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{3}}}{\left( {{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}} \right)_{\rm{n}}}\) contains \(10.46\,\% \,\,{\rm{Br}}\) by mass; the value of n is
1 54
2 65
3 45
4 35
Explanation:
10.46 g Br is present in 100 g compound \(3 \times 80 \mathrm{~g}\) ' Br ' is present in? \( = \frac{{3 \times 80 \times 100}}{{10.46}} = 2294.45\,\,{\text{gram}}\) \(\therefore\) Mass of \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)_{\text {n }}\) \(=2294.45-[(3 \times 80)+(6 \times 12)+(3 \times 1)]\) \(=1979.45\) \(\therefore \mathrm{n}=\dfrac{1979.45}{44}=45\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306647
A sample of pure compound contains \(\mathrm{1.15 \mathrm{~g}}\) of sodium, \(\mathrm{3.01 \times 10^{22}}\) atoms of carbon and \(\mathrm{0.1 \mathrm{~mol}}\) of oxygen atoms. Its empirical formula is
1 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{3}}\)
2 \(\mathrm{\mathrm{NaCO}_{2}}\)
3 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}}\)
4 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{2}}\)
Explanation:
Moles of \(\mathrm{\mathrm{Na}=\dfrac{1.15 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{C}=\dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{O}=0.1}\) Moles ratio, \(\mathrm{\mathrm{Na}: \mathrm{C}: \mathrm{O}=1: 1: 2}\) \(\mathrm{\therefore}\) Empirical formula \(\mathrm{-\mathrm{NaCO}_{2}}\)
306643
What is the molecular mass of a substance which contains 9 carbon atoms,13 hydrogen atoms and \({\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) g of other component?
1 135
2 108
3 13
4 144
Explanation:
The molecule has C, H and other component. Mass of 9 C atoms = 12 \( \times \) 9 = 108 amu Mass of 13 H atoms = 13 \( \times \) 1 = 13 amu Mass of other component \({\rm{ = }}\frac{{{\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}}}{{{\rm{1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}}}{\rm{ = 14}}{\rm{.04 = 135}}{\rm{.04}}\,\,{\rm{amu}}\) \(\therefore \) Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu \(\therefore \) Mol. mass of substance = 135.04
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306644
A compound contains Nitrogen and hydrogen. Its elemental analysis gave 87.5% N. The correct formula of the compound based on the given data is
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
3 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
4 \({\rm{NH}}\)
Explanation:
Hence, the formula of the compound is \({\rm{N}}{{\rm{H}}_{\rm{2}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306645
\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is
306646
A compound having the formula \({\rm{B}}{{\rm{r}}_{\rm{3}}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{3}}}{\left( {{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}} \right)_{\rm{n}}}\) contains \(10.46\,\% \,\,{\rm{Br}}\) by mass; the value of n is
1 54
2 65
3 45
4 35
Explanation:
10.46 g Br is present in 100 g compound \(3 \times 80 \mathrm{~g}\) ' Br ' is present in? \( = \frac{{3 \times 80 \times 100}}{{10.46}} = 2294.45\,\,{\text{gram}}\) \(\therefore\) Mass of \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)_{\text {n }}\) \(=2294.45-[(3 \times 80)+(6 \times 12)+(3 \times 1)]\) \(=1979.45\) \(\therefore \mathrm{n}=\dfrac{1979.45}{44}=45\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306647
A sample of pure compound contains \(\mathrm{1.15 \mathrm{~g}}\) of sodium, \(\mathrm{3.01 \times 10^{22}}\) atoms of carbon and \(\mathrm{0.1 \mathrm{~mol}}\) of oxygen atoms. Its empirical formula is
1 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{3}}\)
2 \(\mathrm{\mathrm{NaCO}_{2}}\)
3 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}}\)
4 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{2}}\)
Explanation:
Moles of \(\mathrm{\mathrm{Na}=\dfrac{1.15 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{C}=\dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{O}=0.1}\) Moles ratio, \(\mathrm{\mathrm{Na}: \mathrm{C}: \mathrm{O}=1: 1: 2}\) \(\mathrm{\therefore}\) Empirical formula \(\mathrm{-\mathrm{NaCO}_{2}}\)
306643
What is the molecular mass of a substance which contains 9 carbon atoms,13 hydrogen atoms and \({\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) g of other component?
1 135
2 108
3 13
4 144
Explanation:
The molecule has C, H and other component. Mass of 9 C atoms = 12 \( \times \) 9 = 108 amu Mass of 13 H atoms = 13 \( \times \) 1 = 13 amu Mass of other component \({\rm{ = }}\frac{{{\rm{2}}{\rm{.33 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}}}{{{\rm{1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 24}}}}}}{\rm{ = 14}}{\rm{.04 = 135}}{\rm{.04}}\,\,{\rm{amu}}\) \(\therefore \) Total mass of one molecule = 108 + 13 + 14.04 = 135.04 amu \(\therefore \) Mol. mass of substance = 135.04
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306644
A compound contains Nitrogen and hydrogen. Its elemental analysis gave 87.5% N. The correct formula of the compound based on the given data is
1 \({\rm{N}}{{\rm{H}}_{\rm{2}}}\)
2 \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)
3 \({{\rm{N}}_{\rm{3}}}{\rm{H}}\)
4 \({\rm{NH}}\)
Explanation:
Hence, the formula of the compound is \({\rm{N}}{{\rm{H}}_{\rm{2}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306645
\(60 \mathrm{~g}\) of a compound on analysis produced \(24 \mathrm{~g}\) carbon, \(4 \mathrm{~g}\) hydrogen and \(32 \mathrm{~g}\) oxygen. The empirical formula of the compound is
306646
A compound having the formula \({\rm{B}}{{\rm{r}}_{\rm{3}}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{3}}}{\left( {{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}} \right)_{\rm{n}}}\) contains \(10.46\,\% \,\,{\rm{Br}}\) by mass; the value of n is
1 54
2 65
3 45
4 35
Explanation:
10.46 g Br is present in 100 g compound \(3 \times 80 \mathrm{~g}\) ' Br ' is present in? \( = \frac{{3 \times 80 \times 100}}{{10.46}} = 2294.45\,\,{\text{gram}}\) \(\therefore\) Mass of \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)_{\text {n }}\) \(=2294.45-[(3 \times 80)+(6 \times 12)+(3 \times 1)]\) \(=1979.45\) \(\therefore \mathrm{n}=\dfrac{1979.45}{44}=45\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306647
A sample of pure compound contains \(\mathrm{1.15 \mathrm{~g}}\) of sodium, \(\mathrm{3.01 \times 10^{22}}\) atoms of carbon and \(\mathrm{0.1 \mathrm{~mol}}\) of oxygen atoms. Its empirical formula is
1 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{3}}\)
2 \(\mathrm{\mathrm{NaCO}_{2}}\)
3 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}}\)
4 \(\mathrm{\mathrm{Na}_{2} \mathrm{CO}_{2}}\)
Explanation:
Moles of \(\mathrm{\mathrm{Na}=\dfrac{1.15 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{C}=\dfrac{3.01 \times 10^{22}}{6.02 \times 10^{23}}=0.05 \mathrm{~mol}}\) Moles of \(\mathrm{\mathrm{O}=0.1}\) Moles ratio, \(\mathrm{\mathrm{Na}: \mathrm{C}: \mathrm{O}=1: 1: 2}\) \(\mathrm{\therefore}\) Empirical formula \(\mathrm{-\mathrm{NaCO}_{2}}\)
