306747
The ratio masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
1 \(1:4\)
2 \(1:8\)
3 \(7:32\)
4 \(3:16\)
Explanation:
The number of moles is given by \({\rm{n = }}\frac{{{\rm{weight}}\,\,{\rm{(W)}}}}{{{\rm{molecular}}{\mkern 1mu} {\mkern 1mu} {\rm{weight}}\,\,{\rm{(M)}}}}\) Thus, ratio of moles of \({{\rm{O}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{N}}_{\rm{2}}}\) is given by \(\frac{{{{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}}}\left( {\frac{{\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}}}{{\frac{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}} \right)\) \({\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{28}}}}{{{\rm{32}}}}} \right){\rm{ = }}\frac{{\rm{7}}}{{{\rm{32}}}}\) Hence, ratio of \({{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{is}}{\mkern 1mu} {\mkern 1mu} {\rm{7:32}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306748
The number of atoms present in 4.25 grams of \(\mathrm{\mathrm{NH}_{3}}\) is approximately.
1 \(\mathrm{1 \times 10^{23}}\)
2 \(\mathrm{8 \times 10^{20}}\)
3 \(\mathrm{2 \times 10^{23}}\)
4 \(\mathrm{6.02 \times 10^{23}}\)
Explanation:
Molar mass of ammonia \(\mathrm{=17 \mathrm{~g} /}\) mole No. of moles \(\mathrm{=(4.25 / 17)=0.25}\) moles No. of atoms \(\mathrm{=4 \times 0.25 \times 6.02 \times 10^{23}}\) \(\rm{=6.02 \times 10^{23}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306749
In which one of the following, does the given amount of chlorine exert the least pressure in a vessel of capacity \({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{d}}{{\rm{m}}^{\rm{3}}}\) at 273 K?
306751
If \(\mathrm{224 \mathrm{ml}}\). of a triatomic gas has a mass of \(\mathrm{1 \mathrm{~g}}\) at \(\mathrm{273 \mathrm{~K}}\) and \(\mathrm{1 \mathrm{~atm}}\) pressure, then the mass of one atom is
1 \(\mathrm{8.30 \times 10^{-23} \mathrm{~g}}\)
2 \(\mathrm{6.24 \times 10^{-23} \mathrm{~g}}\)
3 \(\mathrm{2.08 \times 10^{-23} \mathrm{~g}}\)
4 \(\mathrm{5.54 \times 10^{-23} \mathrm{~g}}\)
Explanation:
No. of moles of triatomic gas \({\text{ = }}\frac{{{\text{224}}\,\,{\text{ml}}}}{{{\text{22400}}\,\,{\text{ml}}\,{\text{mol}}{{\text{e}}^{{\text{ - 1}}}}}}{\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}\) No. of molecules \({\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{ mol}} \times \left( {{\text{6}}{\text{.02}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{molecules}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)\) \(\mathrm{=6.02 \times 10^{21}}\) no. of atoms of gas \(\mathrm{=3 \times 6.02 \times 10^{21}=18.06 \times 10^{21}=1.806 \times 10^{22}}\) \(\mathrm{1.806 \times 10^{22}}\) atoms has mass \(\mathrm{=1 \mathrm{~g}}\) \(\mathrm{\therefore 1}\) atom mass \(\mathrm{=\dfrac{1}{1.806 \times 10^{22} \mathrm{~g}}}\) \(\mathrm{=\dfrac{10^{-22} \mathrm{~g}}{1.806}}\) \(\mathrm{=0.554 \times 10^{-22} \mathrm{~g}}\) \(\mathrm{=5.54 \times 10^{-23} \mathrm{~g}}\)
306747
The ratio masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
1 \(1:4\)
2 \(1:8\)
3 \(7:32\)
4 \(3:16\)
Explanation:
The number of moles is given by \({\rm{n = }}\frac{{{\rm{weight}}\,\,{\rm{(W)}}}}{{{\rm{molecular}}{\mkern 1mu} {\mkern 1mu} {\rm{weight}}\,\,{\rm{(M)}}}}\) Thus, ratio of moles of \({{\rm{O}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{N}}_{\rm{2}}}\) is given by \(\frac{{{{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}}}\left( {\frac{{\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}}}{{\frac{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}} \right)\) \({\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{28}}}}{{{\rm{32}}}}} \right){\rm{ = }}\frac{{\rm{7}}}{{{\rm{32}}}}\) Hence, ratio of \({{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{is}}{\mkern 1mu} {\mkern 1mu} {\rm{7:32}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306748
The number of atoms present in 4.25 grams of \(\mathrm{\mathrm{NH}_{3}}\) is approximately.
1 \(\mathrm{1 \times 10^{23}}\)
2 \(\mathrm{8 \times 10^{20}}\)
3 \(\mathrm{2 \times 10^{23}}\)
4 \(\mathrm{6.02 \times 10^{23}}\)
Explanation:
Molar mass of ammonia \(\mathrm{=17 \mathrm{~g} /}\) mole No. of moles \(\mathrm{=(4.25 / 17)=0.25}\) moles No. of atoms \(\mathrm{=4 \times 0.25 \times 6.02 \times 10^{23}}\) \(\rm{=6.02 \times 10^{23}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306749
In which one of the following, does the given amount of chlorine exert the least pressure in a vessel of capacity \({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{d}}{{\rm{m}}^{\rm{3}}}\) at 273 K?
306751
If \(\mathrm{224 \mathrm{ml}}\). of a triatomic gas has a mass of \(\mathrm{1 \mathrm{~g}}\) at \(\mathrm{273 \mathrm{~K}}\) and \(\mathrm{1 \mathrm{~atm}}\) pressure, then the mass of one atom is
1 \(\mathrm{8.30 \times 10^{-23} \mathrm{~g}}\)
2 \(\mathrm{6.24 \times 10^{-23} \mathrm{~g}}\)
3 \(\mathrm{2.08 \times 10^{-23} \mathrm{~g}}\)
4 \(\mathrm{5.54 \times 10^{-23} \mathrm{~g}}\)
Explanation:
No. of moles of triatomic gas \({\text{ = }}\frac{{{\text{224}}\,\,{\text{ml}}}}{{{\text{22400}}\,\,{\text{ml}}\,{\text{mol}}{{\text{e}}^{{\text{ - 1}}}}}}{\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}\) No. of molecules \({\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{ mol}} \times \left( {{\text{6}}{\text{.02}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{molecules}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)\) \(\mathrm{=6.02 \times 10^{21}}\) no. of atoms of gas \(\mathrm{=3 \times 6.02 \times 10^{21}=18.06 \times 10^{21}=1.806 \times 10^{22}}\) \(\mathrm{1.806 \times 10^{22}}\) atoms has mass \(\mathrm{=1 \mathrm{~g}}\) \(\mathrm{\therefore 1}\) atom mass \(\mathrm{=\dfrac{1}{1.806 \times 10^{22} \mathrm{~g}}}\) \(\mathrm{=\dfrac{10^{-22} \mathrm{~g}}{1.806}}\) \(\mathrm{=0.554 \times 10^{-22} \mathrm{~g}}\) \(\mathrm{=5.54 \times 10^{-23} \mathrm{~g}}\)
306747
The ratio masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
1 \(1:4\)
2 \(1:8\)
3 \(7:32\)
4 \(3:16\)
Explanation:
The number of moles is given by \({\rm{n = }}\frac{{{\rm{weight}}\,\,{\rm{(W)}}}}{{{\rm{molecular}}{\mkern 1mu} {\mkern 1mu} {\rm{weight}}\,\,{\rm{(M)}}}}\) Thus, ratio of moles of \({{\rm{O}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{N}}_{\rm{2}}}\) is given by \(\frac{{{{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}}}\left( {\frac{{\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}}}{{\frac{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}} \right)\) \({\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{28}}}}{{{\rm{32}}}}} \right){\rm{ = }}\frac{{\rm{7}}}{{{\rm{32}}}}\) Hence, ratio of \({{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{is}}{\mkern 1mu} {\mkern 1mu} {\rm{7:32}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306748
The number of atoms present in 4.25 grams of \(\mathrm{\mathrm{NH}_{3}}\) is approximately.
1 \(\mathrm{1 \times 10^{23}}\)
2 \(\mathrm{8 \times 10^{20}}\)
3 \(\mathrm{2 \times 10^{23}}\)
4 \(\mathrm{6.02 \times 10^{23}}\)
Explanation:
Molar mass of ammonia \(\mathrm{=17 \mathrm{~g} /}\) mole No. of moles \(\mathrm{=(4.25 / 17)=0.25}\) moles No. of atoms \(\mathrm{=4 \times 0.25 \times 6.02 \times 10^{23}}\) \(\rm{=6.02 \times 10^{23}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306749
In which one of the following, does the given amount of chlorine exert the least pressure in a vessel of capacity \({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{d}}{{\rm{m}}^{\rm{3}}}\) at 273 K?
306751
If \(\mathrm{224 \mathrm{ml}}\). of a triatomic gas has a mass of \(\mathrm{1 \mathrm{~g}}\) at \(\mathrm{273 \mathrm{~K}}\) and \(\mathrm{1 \mathrm{~atm}}\) pressure, then the mass of one atom is
1 \(\mathrm{8.30 \times 10^{-23} \mathrm{~g}}\)
2 \(\mathrm{6.24 \times 10^{-23} \mathrm{~g}}\)
3 \(\mathrm{2.08 \times 10^{-23} \mathrm{~g}}\)
4 \(\mathrm{5.54 \times 10^{-23} \mathrm{~g}}\)
Explanation:
No. of moles of triatomic gas \({\text{ = }}\frac{{{\text{224}}\,\,{\text{ml}}}}{{{\text{22400}}\,\,{\text{ml}}\,{\text{mol}}{{\text{e}}^{{\text{ - 1}}}}}}{\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}\) No. of molecules \({\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{ mol}} \times \left( {{\text{6}}{\text{.02}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{molecules}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)\) \(\mathrm{=6.02 \times 10^{21}}\) no. of atoms of gas \(\mathrm{=3 \times 6.02 \times 10^{21}=18.06 \times 10^{21}=1.806 \times 10^{22}}\) \(\mathrm{1.806 \times 10^{22}}\) atoms has mass \(\mathrm{=1 \mathrm{~g}}\) \(\mathrm{\therefore 1}\) atom mass \(\mathrm{=\dfrac{1}{1.806 \times 10^{22} \mathrm{~g}}}\) \(\mathrm{=\dfrac{10^{-22} \mathrm{~g}}{1.806}}\) \(\mathrm{=0.554 \times 10^{-22} \mathrm{~g}}\) \(\mathrm{=5.54 \times 10^{-23} \mathrm{~g}}\)
306747
The ratio masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
1 \(1:4\)
2 \(1:8\)
3 \(7:32\)
4 \(3:16\)
Explanation:
The number of moles is given by \({\rm{n = }}\frac{{{\rm{weight}}\,\,{\rm{(W)}}}}{{{\rm{molecular}}{\mkern 1mu} {\mkern 1mu} {\rm{weight}}\,\,{\rm{(M)}}}}\) Thus, ratio of moles of \({{\rm{O}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{N}}_{\rm{2}}}\) is given by \(\frac{{{{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}}}\left( {\frac{{\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}}}{{\frac{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}} \right)\) \({\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{28}}}}{{{\rm{32}}}}} \right){\rm{ = }}\frac{{\rm{7}}}{{{\rm{32}}}}\) Hence, ratio of \({{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{is}}{\mkern 1mu} {\mkern 1mu} {\rm{7:32}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306748
The number of atoms present in 4.25 grams of \(\mathrm{\mathrm{NH}_{3}}\) is approximately.
1 \(\mathrm{1 \times 10^{23}}\)
2 \(\mathrm{8 \times 10^{20}}\)
3 \(\mathrm{2 \times 10^{23}}\)
4 \(\mathrm{6.02 \times 10^{23}}\)
Explanation:
Molar mass of ammonia \(\mathrm{=17 \mathrm{~g} /}\) mole No. of moles \(\mathrm{=(4.25 / 17)=0.25}\) moles No. of atoms \(\mathrm{=4 \times 0.25 \times 6.02 \times 10^{23}}\) \(\rm{=6.02 \times 10^{23}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306749
In which one of the following, does the given amount of chlorine exert the least pressure in a vessel of capacity \({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{d}}{{\rm{m}}^{\rm{3}}}\) at 273 K?
306751
If \(\mathrm{224 \mathrm{ml}}\). of a triatomic gas has a mass of \(\mathrm{1 \mathrm{~g}}\) at \(\mathrm{273 \mathrm{~K}}\) and \(\mathrm{1 \mathrm{~atm}}\) pressure, then the mass of one atom is
1 \(\mathrm{8.30 \times 10^{-23} \mathrm{~g}}\)
2 \(\mathrm{6.24 \times 10^{-23} \mathrm{~g}}\)
3 \(\mathrm{2.08 \times 10^{-23} \mathrm{~g}}\)
4 \(\mathrm{5.54 \times 10^{-23} \mathrm{~g}}\)
Explanation:
No. of moles of triatomic gas \({\text{ = }}\frac{{{\text{224}}\,\,{\text{ml}}}}{{{\text{22400}}\,\,{\text{ml}}\,{\text{mol}}{{\text{e}}^{{\text{ - 1}}}}}}{\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}\) No. of molecules \({\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{ mol}} \times \left( {{\text{6}}{\text{.02}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{molecules}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)\) \(\mathrm{=6.02 \times 10^{21}}\) no. of atoms of gas \(\mathrm{=3 \times 6.02 \times 10^{21}=18.06 \times 10^{21}=1.806 \times 10^{22}}\) \(\mathrm{1.806 \times 10^{22}}\) atoms has mass \(\mathrm{=1 \mathrm{~g}}\) \(\mathrm{\therefore 1}\) atom mass \(\mathrm{=\dfrac{1}{1.806 \times 10^{22} \mathrm{~g}}}\) \(\mathrm{=\dfrac{10^{-22} \mathrm{~g}}{1.806}}\) \(\mathrm{=0.554 \times 10^{-22} \mathrm{~g}}\) \(\mathrm{=5.54 \times 10^{-23} \mathrm{~g}}\)
306747
The ratio masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is
1 \(1:4\)
2 \(1:8\)
3 \(7:32\)
4 \(3:16\)
Explanation:
The number of moles is given by \({\rm{n = }}\frac{{{\rm{weight}}\,\,{\rm{(W)}}}}{{{\rm{molecular}}{\mkern 1mu} {\mkern 1mu} {\rm{weight}}\,\,{\rm{(M)}}}}\) Thus, ratio of moles of \({{\rm{O}}_{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{N}}_{\rm{2}}}\) is given by \(\frac{{{{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}}}\left( {\frac{{\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}}}{{\frac{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{W}}_{{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{W}}_{{{\rm{N}}_{\rm{2}}}}}}}} \right){\rm{ = }}\left( {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{O}}_{\rm{2}}}}}}}} \right)\) \({\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{4}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{28}}}}{{{\rm{32}}}}} \right){\rm{ = }}\frac{{\rm{7}}}{{{\rm{32}}}}\) Hence, ratio of \({{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{and}}{\mkern 1mu} {\mkern 1mu} {{\rm{n}}_{{{\rm{N}}_{\rm{2}}}}}{\mkern 1mu} {\mkern 1mu} {\rm{is}}{\mkern 1mu} {\mkern 1mu} {\rm{7:32}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306748
The number of atoms present in 4.25 grams of \(\mathrm{\mathrm{NH}_{3}}\) is approximately.
1 \(\mathrm{1 \times 10^{23}}\)
2 \(\mathrm{8 \times 10^{20}}\)
3 \(\mathrm{2 \times 10^{23}}\)
4 \(\mathrm{6.02 \times 10^{23}}\)
Explanation:
Molar mass of ammonia \(\mathrm{=17 \mathrm{~g} /}\) mole No. of moles \(\mathrm{=(4.25 / 17)=0.25}\) moles No. of atoms \(\mathrm{=4 \times 0.25 \times 6.02 \times 10^{23}}\) \(\rm{=6.02 \times 10^{23}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306749
In which one of the following, does the given amount of chlorine exert the least pressure in a vessel of capacity \({\rm{1}}{\mkern 1mu} {\mkern 1mu} {\rm{d}}{{\rm{m}}^{\rm{3}}}\) at 273 K?
306751
If \(\mathrm{224 \mathrm{ml}}\). of a triatomic gas has a mass of \(\mathrm{1 \mathrm{~g}}\) at \(\mathrm{273 \mathrm{~K}}\) and \(\mathrm{1 \mathrm{~atm}}\) pressure, then the mass of one atom is
1 \(\mathrm{8.30 \times 10^{-23} \mathrm{~g}}\)
2 \(\mathrm{6.24 \times 10^{-23} \mathrm{~g}}\)
3 \(\mathrm{2.08 \times 10^{-23} \mathrm{~g}}\)
4 \(\mathrm{5.54 \times 10^{-23} \mathrm{~g}}\)
Explanation:
No. of moles of triatomic gas \({\text{ = }}\frac{{{\text{224}}\,\,{\text{ml}}}}{{{\text{22400}}\,\,{\text{ml}}\,{\text{mol}}{{\text{e}}^{{\text{ - 1}}}}}}{\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{mol}}\) No. of molecules \({\text{ = 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{ mol}} \times \left( {{\text{6}}{\text{.02}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{molecules}}\,\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)\) \(\mathrm{=6.02 \times 10^{21}}\) no. of atoms of gas \(\mathrm{=3 \times 6.02 \times 10^{21}=18.06 \times 10^{21}=1.806 \times 10^{22}}\) \(\mathrm{1.806 \times 10^{22}}\) atoms has mass \(\mathrm{=1 \mathrm{~g}}\) \(\mathrm{\therefore 1}\) atom mass \(\mathrm{=\dfrac{1}{1.806 \times 10^{22} \mathrm{~g}}}\) \(\mathrm{=\dfrac{10^{-22} \mathrm{~g}}{1.806}}\) \(\mathrm{=0.554 \times 10^{-22} \mathrm{~g}}\) \(\mathrm{=5.54 \times 10^{-23} \mathrm{~g}}\)
