1 it occupies \(2.24 \mathrm{~L}\) at NTP
2 it corresponds to \(1 / 2\) mole of \(\mathrm{CO}\)
3 it corresponds to same mole of \(\mathrm{CO}\) and nitrogen gas
4 it corresponds to \(3.01 \times 10^{23}\) molecules of \(\mathrm{CO}\)
Explanation:
(1) 28 g of CO occupies
\( = 22.4\;\,{\rm{L}}\)\,of CO at NTP
\(\therefore \) 14 g of CO occupies
\( = \frac{{22.4}}{{28}} \times 14\;{\rm{L}}\,{\rm{of}}\,{\rm{CO}}\,{\rm{at}}\,{\rm{NTP}}\)
\( = 11.2\;{\rm{L}}\,{\rm{of}}\,{\rm{CO}}\,{\rm{at}}\,{\rm{NTP}}\)
Thus 14 g of CO occupies 11.2L volume at NTP
(2) Number of moles of CO
\({\rm{ = }}\frac{{{\rm{mass}}\,{\rm{of}}\,{\rm{CO}}}}{{{\rm{molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{CO}}}}\)
\({\rm{ = }}\frac{{{\rm{14}}\;{\rm{g}}}}{{{\rm{28}}\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\) mole of CO
Thus 14 g CO equal to \(\frac{1}{2}\) mole of CO.
(3)As the molar mass of CO and \({{\rm{N}}_2}\) gas are the same ,thus 14g of CO corresponds to same moles of \({{\text{N}}_2}\) gas.
(4) 28 g of CO contains
\( = 6.022 \times {10^{23}}{\rm{molecules}}\)
\(\therefore \) 14 g of CO contains
\( = \frac{{6.022 \times {{10}^{23}}}}{{28}} \times 14\)
\( = 3.011 \times {10^{23}}{\rm{molecules}}\)
Thus,14 g of CO corresponds to \(3.011 \times {10^{23}}{\rm{molecules}}\).
Hence, option (1) is wrong, while rest of the options are correct.