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ICSE Class 10 Chemistry Question 50 of 72

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 50

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Question

Question 47(c)

A gas cylinder contains 24 × 1024 molecules of nitrogen gas. If Avogadro's number is 6 × 1023 and the relative atomic mass of nitrogen is 14, calculate :

(1) Mass of nitrogen gas in the cylinder.

(2) Volume of nitrogen at STP in dm3

Answer

(1) Gram molecular mass of N2 = 2(14) = 28 g

6 × 1023 molecules of N2 weighs 28 g

∴ 24 × 1024 molecules will weigh 286×1023\dfrac{28}{6 \times 10^{23}} x 24 × 1024 = 1120 g

(2) 6 × 1023 molecules of N2 occupies 22.4 dm3

∴ 24 × 1024 molecules of N2 will occupy 22.46×1023\dfrac{22.4}{6 \times 10^{23}} x 24 × 1024 = 896 dm3