| Column I | Column II |
| (a) 28 g of He | (i) 58.5 u |
| (b) 0.5 mole of \[{{\text{O}}_{2}}\] | (ii) 7 Mol |
| (c) Molecular mass of common salt | (iii) 60 g |
| (d) 1.5 mole of \[\text{Ca}\] | (iv) 16 g |
A) (a) - (iv), (b) - (ii), (c) - (iii), (d) - (i)
B) (a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)
C) (a) - (ii), (b) - (i), (c) - (iii), (d) - (iv)
D) (a) - (ii), (b) - (iv), (c) - (iii), (d) - (i)
Correct Answer: B
Solution :
(a) 4 g of He = 1 mole of He \[\therefore \] 28 g of \[\text{He=}\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ 28}\] moles of He =7 moles of He (b) 1 mole of \[{{\text{O}}_{\text{2}}}\text{ = 32 g of }{{\text{O}}_{\text{2}}}\] 0.5 mole of \[{{\text{O}}_{\text{2}}}\text{=32}\times \text{0}\text{.5 g}\]of\[{{\text{O}}_{\text{2}}}=16g\] of \[{{\text{O}}_{\text{2}}}\] (c) Molecular mass of common salt (NaCI) =23 + 35.5 = 58.5 u (d) 1 mole of Ca = 40 g of Ca 1.5 mole of \[\text{Ca = 40}\times \text{1}\text{.5 g}\] of Ca = 60 g of Ca
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