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Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87°C and 380 mm Hg pressure. (1 litre of hydrogen at s.t.p. weighs 0.09 g)

Stoichiometry

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Answer

Given,

Initial ConditionsFinal Conditions (s.t.p.)
P1 = 380 mm of HgP2 = 760 mm of Hg
V1 = 360 cm3V2 = x
T1 = 87 + 273 K = 360 KT2 = 273 K

Using the gas equation,

P1V1T1=P2V2T2\dfrac{P{1}V{1}}{T{1}} = \dfrac{P{2}V{2}}{T{2}}

Substituting the values we get,

380×360360=760×x273x=38×27376x=10,37476x=136.5 cm3=0.1365 lit.\dfrac{380 \times 360}{360} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{38 \times 273}{76 } \\[0.5em] x = \dfrac{10,374}{76 } \\[0.5em] x = 136.5 \text{ cm}^3 = 0.1365 \text{ lit.}

At s.t.p. 0.1365 lit weighs 0.546 g

Therefore, 1 lit of gas weighs = 0.5460.1365\dfrac{0.546}{0.1365} x 1 = 4 g

Vapour density of gas =

Wt. of certain volume of gas Wt. of same volume of H2=40.09=44.444\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{4}{0.09} \\[0.5em] = 44.444

Molecular weight = 2 x Vapour density
= 2 x 44.444 = 88.888 = 88.89 g

Hence, relative molecular mass of gas 88.89 g

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