3 moles of oxygen is in 1 mole of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\). 1.5 moles of oxygen is in \(\frac{{\rm{1}}}{{\rm{3}}}{\rm{ \times 1}}{\rm{.5}}\) \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.5}}\) mol of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306776
The weight of \(112 \mathrm{~mL}\) of oxygen at NTP is
1 \(0.64 \mathrm{~g}\)
2 \(0.96 \mathrm{~g}\)
3 \(0.32 \mathrm{~g}\)
4 \(0.16 \mathrm{~g}\)
Explanation:
At NTP/STP \({\text{n}} = \frac{{\text{V}}}{{22400(\;{\text{1mL}})}} = \frac{{\text{w}}}{{\text{M}}}\) \(\therefore {\rm{w}} = \frac{{{\rm{V}} \times {\rm{M}}}}{{22,400}}\) Given,\({\rm{V}} = 112\;{\rm{mL}}\) and molar mass of oxygen\( = 32\;{\rm{g}}/{\rm{mol}}\) \(\therefore \;\;{\mkern 1mu} {\kern 1pt} {\rm{w}} = \frac{{112 \times 32}}{{22400\;{\rm{mL}}}} \Rightarrow {\rm{w}} = 0.16\;{\rm{g}}\) Hence, weight of \(112\;{\rm{mL}}\,\)of oxygen at\({\rm{NTP}}/{\rm{STP}} = 0.16\;{\rm{g}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306777
The number of carbon atoms present in a signature, if a signature written by carbon pencil weights \({\rm{1}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{g}}\) is
306779
Number of atoms in 558.5g Fe \(\left( {at.\,\,wt.\,\,of\,\,Fe = 55.85\,g\,mo{l^{ - 1}}} \right)\) is
1 Twice that in 60 g carbon
2 \(6.023 \times {10^{22}}\)
3 Half that in 8 g He
4 \(558.5 \times 6.023 \times {10^{23}}\)
Explanation:
No.of moles Fe \({\rm{ = }}\frac{{{\rm{558}}{\rm{.5}}}}{{{\rm{55}}{\rm{.85}}}}{\rm{ = 10 moles = 10}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}{\rm{.}}\) No. of moles in 60 g of \({\rm{C = }}\frac{{{\rm{60}}}}{{{\rm{12}}}}{\rm{ = 5}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) \({\rm{ = 5}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}\) Number of atoms in \({\rm{Fe = 2 \times }}\) No. of atoms in C.
3 moles of oxygen is in 1 mole of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\). 1.5 moles of oxygen is in \(\frac{{\rm{1}}}{{\rm{3}}}{\rm{ \times 1}}{\rm{.5}}\) \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.5}}\) mol of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306776
The weight of \(112 \mathrm{~mL}\) of oxygen at NTP is
1 \(0.64 \mathrm{~g}\)
2 \(0.96 \mathrm{~g}\)
3 \(0.32 \mathrm{~g}\)
4 \(0.16 \mathrm{~g}\)
Explanation:
At NTP/STP \({\text{n}} = \frac{{\text{V}}}{{22400(\;{\text{1mL}})}} = \frac{{\text{w}}}{{\text{M}}}\) \(\therefore {\rm{w}} = \frac{{{\rm{V}} \times {\rm{M}}}}{{22,400}}\) Given,\({\rm{V}} = 112\;{\rm{mL}}\) and molar mass of oxygen\( = 32\;{\rm{g}}/{\rm{mol}}\) \(\therefore \;\;{\mkern 1mu} {\kern 1pt} {\rm{w}} = \frac{{112 \times 32}}{{22400\;{\rm{mL}}}} \Rightarrow {\rm{w}} = 0.16\;{\rm{g}}\) Hence, weight of \(112\;{\rm{mL}}\,\)of oxygen at\({\rm{NTP}}/{\rm{STP}} = 0.16\;{\rm{g}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306777
The number of carbon atoms present in a signature, if a signature written by carbon pencil weights \({\rm{1}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{g}}\) is
306779
Number of atoms in 558.5g Fe \(\left( {at.\,\,wt.\,\,of\,\,Fe = 55.85\,g\,mo{l^{ - 1}}} \right)\) is
1 Twice that in 60 g carbon
2 \(6.023 \times {10^{22}}\)
3 Half that in 8 g He
4 \(558.5 \times 6.023 \times {10^{23}}\)
Explanation:
No.of moles Fe \({\rm{ = }}\frac{{{\rm{558}}{\rm{.5}}}}{{{\rm{55}}{\rm{.85}}}}{\rm{ = 10 moles = 10}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}{\rm{.}}\) No. of moles in 60 g of \({\rm{C = }}\frac{{{\rm{60}}}}{{{\rm{12}}}}{\rm{ = 5}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) \({\rm{ = 5}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}\) Number of atoms in \({\rm{Fe = 2 \times }}\) No. of atoms in C.
3 moles of oxygen is in 1 mole of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\). 1.5 moles of oxygen is in \(\frac{{\rm{1}}}{{\rm{3}}}{\rm{ \times 1}}{\rm{.5}}\) \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.5}}\) mol of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306776
The weight of \(112 \mathrm{~mL}\) of oxygen at NTP is
1 \(0.64 \mathrm{~g}\)
2 \(0.96 \mathrm{~g}\)
3 \(0.32 \mathrm{~g}\)
4 \(0.16 \mathrm{~g}\)
Explanation:
At NTP/STP \({\text{n}} = \frac{{\text{V}}}{{22400(\;{\text{1mL}})}} = \frac{{\text{w}}}{{\text{M}}}\) \(\therefore {\rm{w}} = \frac{{{\rm{V}} \times {\rm{M}}}}{{22,400}}\) Given,\({\rm{V}} = 112\;{\rm{mL}}\) and molar mass of oxygen\( = 32\;{\rm{g}}/{\rm{mol}}\) \(\therefore \;\;{\mkern 1mu} {\kern 1pt} {\rm{w}} = \frac{{112 \times 32}}{{22400\;{\rm{mL}}}} \Rightarrow {\rm{w}} = 0.16\;{\rm{g}}\) Hence, weight of \(112\;{\rm{mL}}\,\)of oxygen at\({\rm{NTP}}/{\rm{STP}} = 0.16\;{\rm{g}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306777
The number of carbon atoms present in a signature, if a signature written by carbon pencil weights \({\rm{1}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{g}}\) is
306779
Number of atoms in 558.5g Fe \(\left( {at.\,\,wt.\,\,of\,\,Fe = 55.85\,g\,mo{l^{ - 1}}} \right)\) is
1 Twice that in 60 g carbon
2 \(6.023 \times {10^{22}}\)
3 Half that in 8 g He
4 \(558.5 \times 6.023 \times {10^{23}}\)
Explanation:
No.of moles Fe \({\rm{ = }}\frac{{{\rm{558}}{\rm{.5}}}}{{{\rm{55}}{\rm{.85}}}}{\rm{ = 10 moles = 10}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}{\rm{.}}\) No. of moles in 60 g of \({\rm{C = }}\frac{{{\rm{60}}}}{{{\rm{12}}}}{\rm{ = 5}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) \({\rm{ = 5}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}\) Number of atoms in \({\rm{Fe = 2 \times }}\) No. of atoms in C.
3 moles of oxygen is in 1 mole of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\). 1.5 moles of oxygen is in \(\frac{{\rm{1}}}{{\rm{3}}}{\rm{ \times 1}}{\rm{.5}}\) \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.5}}\) mol of \({\rm{BaC}}{{\rm{O}}_{\rm{3}}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306776
The weight of \(112 \mathrm{~mL}\) of oxygen at NTP is
1 \(0.64 \mathrm{~g}\)
2 \(0.96 \mathrm{~g}\)
3 \(0.32 \mathrm{~g}\)
4 \(0.16 \mathrm{~g}\)
Explanation:
At NTP/STP \({\text{n}} = \frac{{\text{V}}}{{22400(\;{\text{1mL}})}} = \frac{{\text{w}}}{{\text{M}}}\) \(\therefore {\rm{w}} = \frac{{{\rm{V}} \times {\rm{M}}}}{{22,400}}\) Given,\({\rm{V}} = 112\;{\rm{mL}}\) and molar mass of oxygen\( = 32\;{\rm{g}}/{\rm{mol}}\) \(\therefore \;\;{\mkern 1mu} {\kern 1pt} {\rm{w}} = \frac{{112 \times 32}}{{22400\;{\rm{mL}}}} \Rightarrow {\rm{w}} = 0.16\;{\rm{g}}\) Hence, weight of \(112\;{\rm{mL}}\,\)of oxygen at\({\rm{NTP}}/{\rm{STP}} = 0.16\;{\rm{g}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306777
The number of carbon atoms present in a signature, if a signature written by carbon pencil weights \({\rm{1}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{g}}\) is
306779
Number of atoms in 558.5g Fe \(\left( {at.\,\,wt.\,\,of\,\,Fe = 55.85\,g\,mo{l^{ - 1}}} \right)\) is
1 Twice that in 60 g carbon
2 \(6.023 \times {10^{22}}\)
3 Half that in 8 g He
4 \(558.5 \times 6.023 \times {10^{23}}\)
Explanation:
No.of moles Fe \({\rm{ = }}\frac{{{\rm{558}}{\rm{.5}}}}{{{\rm{55}}{\rm{.85}}}}{\rm{ = 10 moles = 10}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}{\rm{.}}\) No. of moles in 60 g of \({\rm{C = }}\frac{{{\rm{60}}}}{{{\rm{12}}}}{\rm{ = 5}}{\mkern 1mu} {\mkern 1mu} {\rm{moles}}\) \({\rm{ = 5}}{{\rm{N}}_{\rm{A}}}{\mkern 1mu} {\mkern 1mu} {\rm{atoms}}\) Number of atoms in \({\rm{Fe = 2 \times }}\) No. of atoms in C.
