306713
Calculate the number of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) and \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) ions in 222 g anhydrous \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)
Molecular wt. of \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}{\rm{ = 111g}}\) \(\because {\text{111}}\,\,{\text{g}}\,\,{\text{CaC}}{{\text{l}}_{\text{2}}}\) has \({{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}{\rm{ = 2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) Also \({\mkern 1mu} {\mkern 1mu} {\rm{111}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \({\rm{2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}{\rm{ = 4}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306714
The number of sodium atoms in 2 moles of sodium ferrocyanide is
1 \(12 \times 10^{23}\)
2 \(26 \times 10^{23}\)
3 \(34 \times 10^{23}\)
4 \(48 \times 10^{23}\)
Explanation:
The formula of sodium ferrocyanide is \({\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{({\rm{CN}})}_6}} \right]\) suggests that it has four sodium atoms. Number of \({\rm{Na}}\)-atoms =number of moles \( \times \)number of atoms per molecule\( \times \)Avagadro's number \( = 2 \times 4 \times 6.023 \times {10^{23}}\,{\rm{Na - atoms}}\) \({\rm{ = }}\,{\rm{48}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Na - atoms}}\) \( \sim 48 \times {10^{23}}{\text{Na - atoms}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306715
Which of the following will not have a mass of \(\mathrm{10 \mathrm{~g}}\) ?
306713
Calculate the number of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) and \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) ions in 222 g anhydrous \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)
Molecular wt. of \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}{\rm{ = 111g}}\) \(\because {\text{111}}\,\,{\text{g}}\,\,{\text{CaC}}{{\text{l}}_{\text{2}}}\) has \({{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}{\rm{ = 2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) Also \({\mkern 1mu} {\mkern 1mu} {\rm{111}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \({\rm{2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}{\rm{ = 4}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306714
The number of sodium atoms in 2 moles of sodium ferrocyanide is
1 \(12 \times 10^{23}\)
2 \(26 \times 10^{23}\)
3 \(34 \times 10^{23}\)
4 \(48 \times 10^{23}\)
Explanation:
The formula of sodium ferrocyanide is \({\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{({\rm{CN}})}_6}} \right]\) suggests that it has four sodium atoms. Number of \({\rm{Na}}\)-atoms =number of moles \( \times \)number of atoms per molecule\( \times \)Avagadro's number \( = 2 \times 4 \times 6.023 \times {10^{23}}\,{\rm{Na - atoms}}\) \({\rm{ = }}\,{\rm{48}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Na - atoms}}\) \( \sim 48 \times {10^{23}}{\text{Na - atoms}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306715
Which of the following will not have a mass of \(\mathrm{10 \mathrm{~g}}\) ?
306713
Calculate the number of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) and \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) ions in 222 g anhydrous \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)
Molecular wt. of \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}{\rm{ = 111g}}\) \(\because {\text{111}}\,\,{\text{g}}\,\,{\text{CaC}}{{\text{l}}_{\text{2}}}\) has \({{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}{\rm{ = 2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) Also \({\mkern 1mu} {\mkern 1mu} {\rm{111}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \({\rm{2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}{\rm{ = 4}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306714
The number of sodium atoms in 2 moles of sodium ferrocyanide is
1 \(12 \times 10^{23}\)
2 \(26 \times 10^{23}\)
3 \(34 \times 10^{23}\)
4 \(48 \times 10^{23}\)
Explanation:
The formula of sodium ferrocyanide is \({\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{({\rm{CN}})}_6}} \right]\) suggests that it has four sodium atoms. Number of \({\rm{Na}}\)-atoms =number of moles \( \times \)number of atoms per molecule\( \times \)Avagadro's number \( = 2 \times 4 \times 6.023 \times {10^{23}}\,{\rm{Na - atoms}}\) \({\rm{ = }}\,{\rm{48}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Na - atoms}}\) \( \sim 48 \times {10^{23}}{\text{Na - atoms}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306715
Which of the following will not have a mass of \(\mathrm{10 \mathrm{~g}}\) ?
306713
Calculate the number of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) and \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) ions in 222 g anhydrous \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\)
Molecular wt. of \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}{\rm{ = 111g}}\) \(\because {\text{111}}\,\,{\text{g}}\,\,{\text{CaC}}{{\text{l}}_{\text{2}}}\) has \({{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}\) \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}{\rm{ = 2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{a}}^{{\rm{ + 2}}}}\) Also \({\mkern 1mu} {\mkern 1mu} {\rm{111}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \({\rm{2}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\) \(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{222}}{\mkern 1mu} {\mkern 1mu} {\rm{g}}{\mkern 1mu} {\mkern 1mu} {\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) has \(\frac{{{{\rm{N}}_{\rm{A}}}{\rm{ \times 222}}}}{{{\rm{111}}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}{\rm{ = 4}}{{\rm{N}}_{\rm{A}}}\) ions of \({\rm{C}}{{\rm{l}}^{\rm{ - }}}\).
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306714
The number of sodium atoms in 2 moles of sodium ferrocyanide is
1 \(12 \times 10^{23}\)
2 \(26 \times 10^{23}\)
3 \(34 \times 10^{23}\)
4 \(48 \times 10^{23}\)
Explanation:
The formula of sodium ferrocyanide is \({\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{({\rm{CN}})}_6}} \right]\) suggests that it has four sodium atoms. Number of \({\rm{Na}}\)-atoms =number of moles \( \times \)number of atoms per molecule\( \times \)Avagadro's number \( = 2 \times 4 \times 6.023 \times {10^{23}}\,{\rm{Na - atoms}}\) \({\rm{ = }}\,{\rm{48}}{\rm{.2 \times 1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Na - atoms}}\) \( \sim 48 \times {10^{23}}{\text{Na - atoms}}\)
CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY
306715
Which of the following will not have a mass of \(\mathrm{10 \mathrm{~g}}\) ?
