306752
Calculate the weight of one atom of Ag \(\left( {{\rm{At}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{wt}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{Ag = 108}}} \right)\)
\({{\rm{N}}_{\rm{A}}}\) atoms of Ag weigh 108 gm \(\therefore \) 1 atom of Ag weight \({\rm{ = }}\frac{{{\rm{108}}}}{{{{\rm{N}}_{\rm{A}}}}}\) \({\rm{ = }}\frac{{{\rm{108}}}}{{{\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}}}{\rm{ = 17}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) gm.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306753
16 g of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupies 5.6 litre at STP. Assuming ideal gas nature. The value of x is :
1 1
2 2
3 3
4 None of these
Explanation:
5.6 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by 16 g 22. 4 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by \(\frac{{{\rm{16 \times 22}}{\rm{.4}}}}{{{\rm{5}}{\rm{.6}}}}{\rm{ = 64g}}\) i.e., molar mass of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) \(\therefore \) \({\rm{32 + 16x = 64}}\) \({\rm{x = 2}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306757
The vapour density of a gas is 11.2 . The volume occupied by one gram of the gas at STP is
1 \(1.0 \mathrm{~L}\)
2 \(11.2 \mathrm{~L}\)
3 \(22.4 \mathrm{~L}\)
4 None of these
Explanation:
Here, vapour density of gas \( = 11.2\) \(\therefore \) Molecular weight \( = 2 \times \) vapour density \( = 2 \times 11.2 = 22.4\;{\rm{g}}\) 22.4 g of gas occupied volume at STP \( = 22.4\;{\rm{L}}\) \(\therefore 1\;{\rm{g}}\) gas occupied volume at STP =1L
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306758
The maximum number of molecules is present in
1 \(151\,g\,\,of\,\,{H_2}\,\,gas\,\,at\,\,STP\)
2 \(51\,g\,\,of\,\,{N_2}\,\,gas\,\,at\,\,STP\)
3 \(0.5\,g\,\,of\,\,{H_2}\,\,gas\)
4 \(10\,g\,\,of\,\,{O_2}\,\,gas\)
Explanation:
Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{151}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 6}}{\rm{.74}}\) Moles of \({{\rm{N}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{51}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 2}}{\rm{.27}}\) Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.25}}\) Moles of \({{\rm{O}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{32}}}}{\rm{ = 0}}{\rm{.31}}\) Number of molecules = Number of moles \({\rm{ \times }}{{\rm{N}}_{\rm{A}}}\) The larger the number of moles, the more is the number of molecules.
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306752
Calculate the weight of one atom of Ag \(\left( {{\rm{At}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{wt}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{Ag = 108}}} \right)\)
\({{\rm{N}}_{\rm{A}}}\) atoms of Ag weigh 108 gm \(\therefore \) 1 atom of Ag weight \({\rm{ = }}\frac{{{\rm{108}}}}{{{{\rm{N}}_{\rm{A}}}}}\) \({\rm{ = }}\frac{{{\rm{108}}}}{{{\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}}}{\rm{ = 17}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) gm.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306753
16 g of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupies 5.6 litre at STP. Assuming ideal gas nature. The value of x is :
1 1
2 2
3 3
4 None of these
Explanation:
5.6 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by 16 g 22. 4 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by \(\frac{{{\rm{16 \times 22}}{\rm{.4}}}}{{{\rm{5}}{\rm{.6}}}}{\rm{ = 64g}}\) i.e., molar mass of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) \(\therefore \) \({\rm{32 + 16x = 64}}\) \({\rm{x = 2}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306757
The vapour density of a gas is 11.2 . The volume occupied by one gram of the gas at STP is
1 \(1.0 \mathrm{~L}\)
2 \(11.2 \mathrm{~L}\)
3 \(22.4 \mathrm{~L}\)
4 None of these
Explanation:
Here, vapour density of gas \( = 11.2\) \(\therefore \) Molecular weight \( = 2 \times \) vapour density \( = 2 \times 11.2 = 22.4\;{\rm{g}}\) 22.4 g of gas occupied volume at STP \( = 22.4\;{\rm{L}}\) \(\therefore 1\;{\rm{g}}\) gas occupied volume at STP =1L
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306758
The maximum number of molecules is present in
1 \(151\,g\,\,of\,\,{H_2}\,\,gas\,\,at\,\,STP\)
2 \(51\,g\,\,of\,\,{N_2}\,\,gas\,\,at\,\,STP\)
3 \(0.5\,g\,\,of\,\,{H_2}\,\,gas\)
4 \(10\,g\,\,of\,\,{O_2}\,\,gas\)
Explanation:
Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{151}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 6}}{\rm{.74}}\) Moles of \({{\rm{N}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{51}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 2}}{\rm{.27}}\) Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.25}}\) Moles of \({{\rm{O}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{32}}}}{\rm{ = 0}}{\rm{.31}}\) Number of molecules = Number of moles \({\rm{ \times }}{{\rm{N}}_{\rm{A}}}\) The larger the number of moles, the more is the number of molecules.
306752
Calculate the weight of one atom of Ag \(\left( {{\rm{At}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{wt}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{Ag = 108}}} \right)\)
\({{\rm{N}}_{\rm{A}}}\) atoms of Ag weigh 108 gm \(\therefore \) 1 atom of Ag weight \({\rm{ = }}\frac{{{\rm{108}}}}{{{{\rm{N}}_{\rm{A}}}}}\) \({\rm{ = }}\frac{{{\rm{108}}}}{{{\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}}}{\rm{ = 17}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) gm.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306753
16 g of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupies 5.6 litre at STP. Assuming ideal gas nature. The value of x is :
1 1
2 2
3 3
4 None of these
Explanation:
5.6 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by 16 g 22. 4 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by \(\frac{{{\rm{16 \times 22}}{\rm{.4}}}}{{{\rm{5}}{\rm{.6}}}}{\rm{ = 64g}}\) i.e., molar mass of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) \(\therefore \) \({\rm{32 + 16x = 64}}\) \({\rm{x = 2}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306757
The vapour density of a gas is 11.2 . The volume occupied by one gram of the gas at STP is
1 \(1.0 \mathrm{~L}\)
2 \(11.2 \mathrm{~L}\)
3 \(22.4 \mathrm{~L}\)
4 None of these
Explanation:
Here, vapour density of gas \( = 11.2\) \(\therefore \) Molecular weight \( = 2 \times \) vapour density \( = 2 \times 11.2 = 22.4\;{\rm{g}}\) 22.4 g of gas occupied volume at STP \( = 22.4\;{\rm{L}}\) \(\therefore 1\;{\rm{g}}\) gas occupied volume at STP =1L
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306758
The maximum number of molecules is present in
1 \(151\,g\,\,of\,\,{H_2}\,\,gas\,\,at\,\,STP\)
2 \(51\,g\,\,of\,\,{N_2}\,\,gas\,\,at\,\,STP\)
3 \(0.5\,g\,\,of\,\,{H_2}\,\,gas\)
4 \(10\,g\,\,of\,\,{O_2}\,\,gas\)
Explanation:
Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{151}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 6}}{\rm{.74}}\) Moles of \({{\rm{N}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{51}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 2}}{\rm{.27}}\) Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.25}}\) Moles of \({{\rm{O}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{32}}}}{\rm{ = 0}}{\rm{.31}}\) Number of molecules = Number of moles \({\rm{ \times }}{{\rm{N}}_{\rm{A}}}\) The larger the number of moles, the more is the number of molecules.
306752
Calculate the weight of one atom of Ag \(\left( {{\rm{At}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{wt}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\rm{of}}{\mkern 1mu} {\mkern 1mu} {\rm{Ag = 108}}} \right)\)
\({{\rm{N}}_{\rm{A}}}\) atoms of Ag weigh 108 gm \(\therefore \) 1 atom of Ag weight \({\rm{ = }}\frac{{{\rm{108}}}}{{{{\rm{N}}_{\rm{A}}}}}\) \({\rm{ = }}\frac{{{\rm{108}}}}{{{\rm{6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}}}{\rm{ = 17}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 23}}}}\) gm.
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306753
16 g of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupies 5.6 litre at STP. Assuming ideal gas nature. The value of x is :
1 1
2 2
3 3
4 None of these
Explanation:
5.6 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by 16 g 22. 4 L of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) occupied by \(\frac{{{\rm{16 \times 22}}{\rm{.4}}}}{{{\rm{5}}{\rm{.6}}}}{\rm{ = 64g}}\) i.e., molar mass of \({\rm{S}}{{\rm{O}}_{\rm{x}}}\) \(\therefore \) \({\rm{32 + 16x = 64}}\) \({\rm{x = 2}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306757
The vapour density of a gas is 11.2 . The volume occupied by one gram of the gas at STP is
1 \(1.0 \mathrm{~L}\)
2 \(11.2 \mathrm{~L}\)
3 \(22.4 \mathrm{~L}\)
4 None of these
Explanation:
Here, vapour density of gas \( = 11.2\) \(\therefore \) Molecular weight \( = 2 \times \) vapour density \( = 2 \times 11.2 = 22.4\;{\rm{g}}\) 22.4 g of gas occupied volume at STP \( = 22.4\;{\rm{L}}\) \(\therefore 1\;{\rm{g}}\) gas occupied volume at STP =1L
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306758
The maximum number of molecules is present in
1 \(151\,g\,\,of\,\,{H_2}\,\,gas\,\,at\,\,STP\)
2 \(51\,g\,\,of\,\,{N_2}\,\,gas\,\,at\,\,STP\)
3 \(0.5\,g\,\,of\,\,{H_2}\,\,gas\)
4 \(10\,g\,\,of\,\,{O_2}\,\,gas\)
Explanation:
Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{151}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 6}}{\rm{.74}}\) Moles of \({{\rm{N}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{51}}}}{{{\rm{22}}{\rm{.4}}}}{\rm{ = 2}}{\rm{.27}}\) Moles of \({{\rm{H}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.5}}}}{{\rm{2}}}{\rm{ = 0}}{\rm{.25}}\) Moles of \({{\rm{O}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{10}}}}{{{\rm{32}}}}{\rm{ = 0}}{\rm{.31}}\) Number of molecules = Number of moles \({\rm{ \times }}{{\rm{N}}_{\rm{A}}}\) The larger the number of moles, the more is the number of molecules.