A) It occupies 2.24 L at NTP
B) It corresponds to \[\frac{1}{2}\] mole of CO
C) It corresponds to same mole of CO and \[{{\text{N}}_{\text{2}}}\]
D) It corresponds to \[3.01\times {{10}^{23}}\] molecules of CO
Correct Answer: A
Solution :
Key Idea: 1 mol = molecular mass in gram \[=6.02\times {{10}^{23}}\]molecules/atoms/ions \[=22.4\,L\]at STP (i) Molecular mass of CO = 12 + 16 = 28 28 g of CO = 1 mol \[\therefore \] 14 got CO = 0.5 mol (ii) 1 mole of \[CO=6.023\times {{10}^{23}}\] molecules of CO \[\therefore \] 0.5 mole of \[CO=3.023\text{ }\times {{10}^{23}}\] molecules of CO Molecular mass of \[{{N}_{2}}=14+14=28\] \[\therefore \] \[{{\text{N}}_{\text{2}}}\]and CO will have same moles. \[\therefore \] choice (c) is correct. (iv) \[28\,g(=1\,\text{mol})\]of CO occupies 22.4 L of volume at STP.
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