306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules \(\left( {{\rm{d = }}\,{\rm{0}}.{\rm{75}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right),{\rm{11}}.{\rm{2}}\;{\rm{g}}\) of molecules \(\left( {{\rm{d}}\,{\rm{ = }}\,{\rm{3}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right)\) and 5.6 g of molecules \(({\rm{d}}\,{\rm{ = 1}}.{\rm{5}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}})\). All density measurements are made at STP. Calculate the total number of molecules ( \({\mathrm{N}}\) ) present in the given sample. Report your answer in ' \({\mathrm{10^{23} \mathrm{~N}}}\)'. Assume Avogadro's number as \({\mathrm{6 \times 10^{23}}}\).
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An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
\({\rm{ = 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\mkern 1mu} {\mkern 1mu} {\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\))
306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules \(\left( {{\rm{d = }}\,{\rm{0}}.{\rm{75}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right),{\rm{11}}.{\rm{2}}\;{\rm{g}}\) of molecules \(\left( {{\rm{d}}\,{\rm{ = }}\,{\rm{3}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right)\) and 5.6 g of molecules \(({\rm{d}}\,{\rm{ = 1}}.{\rm{5}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}})\). All density measurements are made at STP. Calculate the total number of molecules ( \({\mathrm{N}}\) ) present in the given sample. Report your answer in ' \({\mathrm{10^{23} \mathrm{~N}}}\)'. Assume Avogadro's number as \({\mathrm{6 \times 10^{23}}}\).
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An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
\({\rm{ = 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\mkern 1mu} {\mkern 1mu} {\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\))
306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules \(\left( {{\rm{d = }}\,{\rm{0}}.{\rm{75}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right),{\rm{11}}.{\rm{2}}\;{\rm{g}}\) of molecules \(\left( {{\rm{d}}\,{\rm{ = }}\,{\rm{3}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right)\) and 5.6 g of molecules \(({\rm{d}}\,{\rm{ = 1}}.{\rm{5}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}})\). All density measurements are made at STP. Calculate the total number of molecules ( \({\mathrm{N}}\) ) present in the given sample. Report your answer in ' \({\mathrm{10^{23} \mathrm{~N}}}\)'. Assume Avogadro's number as \({\mathrm{6 \times 10^{23}}}\).
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An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
\({\rm{ = 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\mkern 1mu} {\mkern 1mu} {\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\))
306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules \(\left( {{\rm{d = }}\,{\rm{0}}.{\rm{75}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right),{\rm{11}}.{\rm{2}}\;{\rm{g}}\) of molecules \(\left( {{\rm{d}}\,{\rm{ = }}\,{\rm{3}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}}} \right)\) and 5.6 g of molecules \(({\rm{d}}\,{\rm{ = 1}}.{\rm{5}}\;{\rm{g\;}}{{\rm{L}}^{{\rm{ - 1}}}})\). All density measurements are made at STP. Calculate the total number of molecules ( \({\mathrm{N}}\) ) present in the given sample. Report your answer in ' \({\mathrm{10^{23} \mathrm{~N}}}\)'. Assume Avogadro's number as \({\mathrm{6 \times 10^{23}}}\).
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An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
\({\rm{ = 6}}{\rm{.023 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\mkern 1mu} {\mkern 1mu} {\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\))