Chemistry Test 71

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Chemistry Test 71

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Chemistry Test 71

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Question 1/2
4 / -1

How many atoms of each of the constituent element are there in 50 gm. of CaCO₃ ?

(at. wt. of Ca = 40, C = 12, O = 16) :-

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A
Ca = 3.01 × 10²³, C = 3.01 × 10²³, O = 9.03 × 10²³

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B
Ca = 9.03 × 10²³, C = 3.01 × 10²³, O = 3.01 × 10²³

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C
Ca = 4.61 × 10²³, C = 3.01 × 10²³, O = 9.03 × 10²³

Correct

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D
Ca = 2.01 × 10²³, C = 2.01 × 10²³, O = 9.03 × 10²³

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Solutions

moles of CaCO₃ = 50/100 = 0.5

molecules of CaCO₃ = 0.5 × N_A = 3.01 × 10²³
Question 2/2
4 / -1

Calculate the volume of H₂S gas (in millilitre) at N.T.P. needed to precipitate cupric sulphide completely from 0.75g of CuCl₂?

[At. wt. of Cu = 63.5, At. wt. of Cl = 35.5]

Reaction: CuCl₂ + H₂S —→ CuS + 2HCl

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A
21.4 ml

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B
14.2 ml

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C
41.2 ml

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D
124.9 ml

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