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ICSE Class 10 Chemistry Question 5 of 18

Solved 2020 Question Paper ICSE Class 10 Chemistry — Question 10

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10
Question

Question 1(f)

Calculate:

(i) The amount of each reactant required to produce 750 ml of carbon dioxide, when two volumes of carbon monoxide combine with one volume of oxygen to produce two volumes of carbon dioxide.

2CO + O2 ⟶ 2CO2

(ii) The volume occupied by 80 g of carbon dioxide at STP.

(iii) Calculate the number of molecules in 4.4 gm of CO2.

[Atomic mass of C= 12, O=16]

(iv) State the law associated in question no. (f)(i) above.

Answer

(i) [By Lussac's law]

2CO+O2⟶2CO22 vol.:1 vol.⟶2 vol.\begin{matrix} 2\text{CO}& + & \text{O}_2 &\longrightarrow & 2\text{CO}_2 \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}

To calculate the amount of CO required :

CO2:CO2 vol.:2 vol.750 ml:x\begin{matrix}\text{CO}_2 & : & \text{CO} & \\ 2 \text{ vol.} & : & 2 \text{ vol.} \\ 750 \text{ ml} & : & \text{x} \end{matrix}

∴x=22×750=750 ml\therefore x = \dfrac{2}{2} \times 750 = 750 \text{ ml}

To calculate the amount of O2 required :

CO2:O2 2 vol.:1 vol.750 cm.3:x\begin{matrix}\text{CO}_2 & : & \text{O}_2 & \\ \text{ 2 vol.} & : & 1 \text{ vol.} \\ 750 \text{ cm.}^3 & : & \text{x} \end{matrix}

∴x=12×750=375 ml\therefore x = \dfrac{1}{2} \times 750 = 375 \text{ ml}

Hence, CO required is 750 ml and O2 required = 375 ml

(ii) Gram molecular mass of CO2 = 12 + 32 = 44 g and occupies 22.4 lit. vol.

∴ 80 g. will occupy = 22.444\dfrac{22.4}{44} x 80 = 40.73 lit.

Hence, volume occupied = 40.73 lits.

(iii) Gram molecular mass of CO2 = 12 + 32 = 44 g and contains 6.023 x 1023 molecules.

∴ 4.4 gm. of CO2 will contain = 6.023×102344\dfrac{6.023 \times 10^{23}}{44} x 4.4 = 6.023 x 1022 molecules.

Hence, number of molecules = 6.023 x 1022 molecules.

(iv) Gay Lussac's law is associated in the question.