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Standard XII

Chemistry

Molarity

44.8 V H2O2...

Question

44.8 V $H_{2} O_{2}$ Solution (500 ml) when exposed to atmosphere looses 11.2 litre of $O_{2}$ at 1 atm & 273K. New molarity of $H_{2} O_{2}$ solution (Assume no change in volume)

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Solution

$2 H_{2} O_{2} ⟶ 2 H_{2} O + O_{2}$
$68 g m ⟶ 22400 m l$ of $O_{2}$
$44.8 g ⟶ ?$
$\Rightarrow \frac{44.8}{68} \times 22400 = 14757.6 m l$ of $O_{2}$
So, $1 m l$ of aqueous $H_{2} O_{2}$ solution produces $\frac{147576}{1000} = 14.757 m l$ of $O_{2}$
Molarity of $H_{2} O_{2} =$ Molecular weight of $H_{2} O_{2} = 34 g m$ , Equivalent mass of $H_{2} O_{2} = \frac{34}{2} = 17$
$68 g m / l$ aq $H_{2} O_{2} = \frac{68}{17} = 4 N$ .

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Similar questions

Which of the following statements are correct for a solution of $H_{2} O_{2}$ having a strength of 17g/litre ?

500

ml of a

H_{2} O_{2}

solution on complete decomposition produces

2

moles of

H_{2} O_{2}

. Calculate the volume strength of

H_{2} O_{2}

solution?

[Assume that the volume of

O_{2}

is measured at

1

atm and

273

Q. The srength of

H_{2} O_{2}

is expressed in several ways like molarity, normality, %(w/V), volume strength, etc. The strength of

10 V

means 1 volume of

H_{2} O_{2}

on decomposition gives 10 volumes of oxygen at 1 atm and

273 K

or 1 litre of

H_{2} O_{2}

gives

10 l i t r e

O_{2}

at 1 atm and

273 K

.
The decomposition of

H_{2} O_{2}

is shown as under:

H_{2} O_{2} (a q) \to H_{2} O (l) + \frac{1}{2} O_{2} (g)

H_{2} O_{2}

can acts as oxidising as well as reducing agent, as oxidisng agent

H_{2} O_{2}

converted into

H_{2} O

and as reducing agent

H_{2} O_{2}

converted into

O_{2}

, both cases n-factor is

2

∴, Normality of H_{2} O_{2} Solution = 2 \times Molarity of H_{2} O_{2} Solution

What is the percentage strength of

11.2 V H_{2} O_{2}

500 mL

of a

H_{2} O_{2}

solution on complete decomposition produces

2

moles of

H_{2} O

, volume strength of

H_{2} O_{2}

solution is:
[Given : Volume of

O_{2}

is measured at

1 atm

and

273 K

]

Q. .............. volume

H_{2} O_{2}

solution means that

1

mL of

H_{2} O_{2}

at NTP gives

1

L of

O_{2}

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44.8 V H2O2 Solution (500 ml) when exposed to atmosphere looses 11.2 litre of O2 at 1 atm & 273K. New molarity of H2O2 solution (Assume no change in volume)

2H2O2⟶2H2O+O268gm⟶22400ml of O244.8g⟶?⇒44.868×22400=14757.6ml of O2So, 1ml of aqueous H2O2 solution produces 1475761000=14.757ml of O2Molarity of H2O2= Molecular weight of H2O2=34gm, Equivalent mass of H2O2=342=1768gm/l aq H2O2=6817=4N .

44.8 V $H_{2} O_{2}$ Solution (500 ml) when exposed to atmosphere looses 11.2 litre of $O_{2}$ at 1 atm & 273K. New molarity of $H_{2} O_{2}$ solution (Assume no change in volume)