Chemistry – 1
1 The oxidation number of carbon in CH₂O is
- 2 +2 0 +4
2 The value of Avogadro constant is
6.022 × 10²³
6.022 × 10²³ mol⁻¹
6.022 × 10²³ atoms
6.022 × 10²³ molecule mol⁻¹
3 The largest number of molecules is in
36 g of water
28 g of carbon monoxide
46 g of ethyl alcohol
54 g of nitrogen pentoxide
4 The oxidation state of nitrogen in N₃H is
- 1/3
- 1
1/2
2
5 The oxidation state of nickel in Ni(CO)₄ is
0 +2 +4 - 4
- 2 +2 0 +4
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We have x + 2(+1) + (–2) = 0. This gives x = 0.2 The value of Avogadro constant is
6.022 × 10²³
6.022 × 10²³ mol⁻¹
6.022 × 10²³ atoms
6.022 × 10²³ molecule mol⁻¹
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answer is 6.022 × 10²³ mol⁻¹3 The largest number of molecules is in
36 g of water
28 g of carbon monoxide
46 g of ethyl alcohol
54 g of nitrogen pentoxide
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Amount of water, M/M = 36 g / 18 g mol⁻¹ = 2 mol ; Amount of carbon monoxide 28 g / 28 g mol⁻¹ = 1 mol ; Amount of ethyl alcohol=46 g / 46 g mol⁻¹ = 1 mol ; Amount of nitrogen pentoxide = 54 g / 108 g mol⁻¹ = 0.5 mol ; Larger the amount of substance, larger the number of molecules. Ans : 36 g of water4 The oxidation state of nitrogen in N₃H is
- 1/3
- 1
1/2
2
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The oxidation of N in N3H may be computed from the expression 3x + 1 = 0. This gives x = - ⅓.5 The oxidation state of nickel in Ni(CO)₄ is
0 +2 +4 - 4
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