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Chemistry

A gas cylinder contains 12 x 1024 molecules of oxygen gas.

Calculate : the mass of O2 present in the cylinder.

(ii) the volume of O2 at S.T.P. present in the cylinder.

[O = 16] Avog. no. is 6 × 1023

Stoichiometry

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Answer

Gram molecular mass of oxygen = 32 g = 1 mole and contains 6 x 1023 molecules i.e., 6 x 1023 molecules weigh 32 g

∴ 12 x 1024 molecules will weigh = 326×1023\dfrac{32}{ 6\times 10^{23}} x 12 x 1024 = 640 g

(ii) 6 x 1023 molecules occupy 22.4 lit. of vol.

∴ 12 x 1024 molecules will occupy = 22.46×1023\dfrac{22.4}{ 6\times 10^{23}} x 12 x 1024 = 448 lit.

Hence, mass of O2 present in the cylinder = 640 g and volume of O2 at S.T.P. present in the cylinder = 448 lit.

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