Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation-2 H 2 O 2 aq - 2 H 2 O l - O 2 g under conditions where 1 mole of gas occupies 24 dm 3 -100 cm 3 of X M solution of H 2 O 2 produces 3 dm 3 of O 2 Thus- X is-A- 2-5B- 1- 0-5D- 0-25

The correct option is A 2.524 dm^3 → 1 mole of O_2 gas. 3 dm^3 →324 = 18 mole of O_2 gas. 2H_2O_2 → 2H_2O + O_2 n_H_2O_2 = 2 × n_O_2 = 2 ×18 = 14moles ∴ X × ...

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Standard XII

Chemistry

Molarity

Hydrogen pero...

Question

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation:
$2 H_{2} O_{2} (a q) \to 2 H_{2} O (l) + O_{2} (g)$
under conditions where $1$ mole of gas occupies $24 d m^{3} . 100 c m^{3}$ of $X M$ solution of $H_{2} O_{2}$ produces $3 d m^{3}$ of $O_{2}$ Thus, $X$ is:

2.5

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Solution

The correct option is A $2.5$
$24 d m^{3} \to 1$ mole of $O_{2}$ gas.
$3 d m^{3} \to \frac{3}{24} = \frac{1}{8}$ mole of $O_{2}$ gas.
$2 H_{2} O_{2} \to 2 H_{2} O + O_{2} n_{H_{2} O_{2}} = 2 \times n_{O_{2}} = 2 \times \frac{1}{8} = \frac{1}{4} m o l e s$
$∴ X \times 0.1 = 0.25 \Rightarrow X = 2.5$

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Similar questions

2 H_{2} O_{2} (a q) ⟶ 2 H_{2} O (l) + O_{2} (g)

Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the given equation under conditions where one mole of gas occupies

24 d m^{3}

100 c m^{3}

x M

solution of

H_{2} O_{2}

produces

3 d m^{3}

O_{2}

Thus,

x

is :

Q. Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen, according to the equation:

2 H_{2} O_{2} (a q) \to 2 H_{2} O (l) + O_{2} (g)

under conditions, where 1 mole of gas occupies

24 d m^{3} . 100 d m^{3}

of X M solution of

H_{2} O_{2}

produces

3 d m^{3}

O_{2}

Thus, X is:

Q. Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation:

2 H_{2} O_{2} (a q) ⟶ 2 H_{2} O (ℓ) + O_{2} (g)

Under conditions, where

1

mole of gas occupies

24 d m^{3}

100 c m^{3}

x M

solution of

H_{2} O_{2}

produces

3 d m^{3}

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. Thus,

x

is:

2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)

100

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H_{2} O_{2}

gives

3

L of

O_{2}

gas under the condition when

1

mol occupies

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x

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Q. The mole fraction of hydrogen peroxide in 30% (weight / weight) aqueous solution of

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is _________________.

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Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen according to the equation: 2H2O2(aq)→2H2O(l)+O2(g) under conditions where 1 mole of gas occupies 24 dm3. 100 cm3 of X M solution of H2O2 produces 3 dm3 of O2 Thus, X is:

The correct option is A 2.524 dm3→1 mole of O2 gas. 3 dm3→324=18 mole of O2 gas. 2H2O2→2H2O+O2nH2O2=2×nO2=2×18=14moles ∴X×0.1=0.25 ⇒X=2.5