1. ⇒ (MHT CET 2023 12th May Evening Shift )

What is the mass of ${KClO}_{3 (s)}$ required to liberate 22. $4 {dm}^{3}$ oxygen at STP during thermal decomposition?

$($ Molar Mass of ${KClO}_{3 (s)} = 122.5 g / mol)$

A. 122.5 g

B. 81.67 g

C. 10.25 g

D. 8.16 g

Correct Option is (B)

MHT CET 2023 12th May Evening Shift Chemistry - States of Matter Question 4 English Explanation

2 moles of ${KClO}_{3} = 2 \times 122.5 = 245 g$

3 moles of $O_{2}$ at STP occupy $= (3 \times 22.4 {dm}^{3})$

Thus, $245 g$ of potassium chlorate will liberate $67.2 {dm}^{3}$ of oxygen gas.

Let ' $x$ ' gram of ${KClO}_{3}$ liberate $22.4 {dm}^{3}$ of oxygen gas at S.T.P.

$∴ x = \frac{245 \times 22.4}{3 \times 22.4} = 81.67 g$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

What volume of ${CO}_{2 (g)}$ at STP is obtained by complete combustion of $6 g$ carbon?

A. $22.4 {dm}^{3}$

B. $11.2 {dm}^{3}$

C. $5.6 {dm}^{3}$

D. $2.24 {dm}^{3}$

Correct Option is (B)

$\begin{aligned} C_{(s)} + O_{2 (g)} ⟶ {CO}_{2 (g)} \\ 1 mol C \equiv 1 mol {CO}_{2 (g)} \\ 6 g C = 0.5 mol C \\ ∴ & 0.5 mol C \equiv 0.5 mol {CO}_{2 (g)} \\ At STP, 1 mol {CO}_{2 (g)} \equiv 22.4 {dm}^{3} \\ ∴ & 0.5 mol {CO}_{2 (g)} = 11.2 {dm}^{3} \end{aligned}$

3. ⇒ (MHT CET 2023 11th May Morning Shift )

What is the volume in ${dm}^{3}$ occupied by $3 mol$ of ammonia gas at STP?

A. 2.24

B. 22.4

C. 56.0

D. 67.2

Correct Option is (D)

$\begin{aligned} Number of moles of a gas (n) \\ = \frac{Volume of a gas at STP}{Molar volume of a gas} \\ ∴ Volume of ammonia gas at STP \\ = Number of moles of the gas (n) \\ \times Molar volume of the gas \\ = 3 mol \times 22.4 {dm}^{3} {mol}^{- 1} \\ = 67.2 {dm}^{3} \end{aligned}$

4. ⇒ (MHT CET 2021 21th September Evening Shift )

What is the volume occupied by 16 g methane gas at STP?

A. 1140 cm $^{3}$

B. 22400 cm $^{3}$

C. 214 cm $^{3}$

D. 12.4 cm $^{3}$

Correct Option is (B)

1 mole of CH $_{4}$ = 16 g of CH $_{4}$ = 22400 cm $^{3}$ at STP

⇒States of Matter