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ICSE Class 10 Chemistry Question 11 of 19

Solved 2017 Question Paper ICSE Class 10 Chemistry — Question 11

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11
Question

Question 5(a)

(i) Propane burns in air according to the following equation :

C3H8 + 5O2 ⟶ 3CO2 + 4H2O.

What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen.

(ii) The mass of 11.2 litres of a certain gas at s.t.p. is 24 g. Find the gram molecular mass of the gas.

Answer

(i) Given, 20% of air contains oxygen

∴ 20% of 1000 cm3 = 20100\dfrac{20}{100} x 1000 = 200 cm3

[By Lussac's law]

C3H8+5O2⟶3CO2+4H2O1 vol.:5 vol.⟶3 vol.:4 vol.\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 &\longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} & : & 4\text{ vol.} \end{matrix}

To calculate the volume of propane consumed :

O2:C3H85 vol.:1 vol.200 cm.3:x\begin{matrix}\text{O}_2 & : & \text{C}_3\text{H}_8 \\ 5 \text{ vol.} & : & 1 \text{ vol.} \\ 200 \text{ cm.}^3 & : & \text{x} \end{matrix}

∴x=15×200=40 cm.3\therefore x = \dfrac{1}{5} \times 200 = 40 \text{ cm.}^3

Hence, volume of propane is consumed is 40 cm3

(ii) 11.2 lit. weighs 24 g

∴ 22.4 lit. will weigh = 2411.2\dfrac{24}{11.2} x 22.4 = 48 g

Hence, gram molecular mass of the gas = 48 g.