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ICSE Class 10 Chemistry Question 22 of 25

Gay Lussac's Law — Avogadro's Law — Mole Concept — Question 22

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Question 22

Calculate the weight of the substance if it's molecular weight is 70 and in the gaseous form occupies 10 lits. at 27°C and 700 mm pressure.

Answer
Initial ConditionsFinal Conditions (s.t.p.)
P1 = 700 mm of HgP2 = 760 mm of Hg
V1 = 10 litV2 = x lit
T1 = 27 + 273 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,

700×10300=760×x273x=700×10×273300×760x=1911228x=8.38 lit\dfrac{700 \times 10}{300} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{700 \times 10 \times 273}{300 \times 760 } \\[0.5em] x = \dfrac{1911}{228 } \\[0.5em] x = 8.38 \text{ lit}

1 gram molecular weight of the gas occupies 22.4 lit. at s.t.p.

∴ 70 g of the gas occupies 22.4 lit. at s.t.p.

If yy g of the gas occupies 8.38 lits, then

y=7022.4×8.38=26.18 gy = \dfrac{70}{22.4} \times 8.38 \\[0.5em] = 26.18 \text{ g}

Hence, weight of substance is 26.18 g