Hydrogen peroxide in aqueous solution decomposes on warming to give oxygen- according to the equation- 2H2O2aq - 2H2Ol - O2g under conditions- where 1 mole of gas occupies 24 dm3 - 100 dm3 of X M solution of H2O2 produces 3 dm3 of O2- Thus- X is-

The correct option is A 2.5 3 dm^3×1 mole24 dm^3=0.125 moles of O_2 2H_2O_2 ⟶ 2H_2O + O_2 2 moles of H_2O_2 produces 1 mole of O_2 . 0. ...

153

You visited us 153 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Potassium Permanganate

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 $24 d m^{3} . 100 d m^{3}$ of X M solution of $H_{2} O_{2}$ produces $3 d m^{3}$ of $O_{2}$ .

Thus, X is:

2.5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 2.5
$3 d m^{3} \times \frac{1 m o l e}{24 d m^{3}} = 0.125 m o l e s o f O_{2}$

$2 H_{2} O_{2} ⟶ 2 H_{2} O + O_{2}$

2 moles of $H_{2} O_{2}$ produces 1 mole of $O_{2}$ .

$0.125$ moles of $O_{2}$ requires $2 \times 0.125$ $H_{2} O_{2} = 0.25$ moles of $H_{2} O_{2}$

Concentration of $H_{2} O_{2} = \frac{0.25}{0.1} = 2.5 M$

Hence, option A is correct.

Suggest Corrections

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) ⟶ 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}

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 c m^{3}

X M

solution of

H_{2} O_{2}

produces

3 d m^{3}

O_{2}

Thus,

X

is:

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

100

mL of

X

molar

H_{2} O_{2}

gives

3

L of

O_{2}

gas under the condition when

1

mol occupies

24

L. The value of

X

is :

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

H_{2} O_{2}

is _________________.

Join BYJU'S Learning Program

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 24dm3.100dm3 of X M solution of H2O2 produces 3dm3 of O2. Thus, X is:

The correct option is A 2.53dm3×1 mole24 dm3=0.125 moles of O22H2O2⟶2H2O+O22 moles of H2O2 produces 1 mole of O2.0.125 moles of O2 requires 2×0.125 H2O2=0.25 moles of H2O2Concentration of H2O2=0.250.1=2.5 MHence, option A is correct.