$34 g$ hydrogen peroxide is present in $1120 m L$ of the solution. The volume strength of $H_{2} O_{2}$ is :

10 V

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

20 V

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

30 V

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

32 V

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

Open in App

Solution

The correct option is A 10 V
$H_{2} O_{2} (a q) \to H_{2} O (l) + \frac{1}{2} O_{2} (g)$
Molecular mass of $H_{2} O_{2} = 34 g m o l^{- 1}$
Moles of $H_{2} O_{2} = \frac{34}{34} = 1 m o l$
Weight of $H_{2} O_{2}$ in $1 m L = \frac{34}{1120} g$
In $1120 m L$ of solution there is $1 m o l$ of $H_{2} O_{2}$
So, in $1000 m L$ of solution, moles of $H_{2} O_{2} = \frac{1 \times 1000}{1120} = \frac{100}{112} m o l$
$1 m o l$ of $H_{2} O_{2}$ gives $11.2 L$ of $O_{2}$ at STP
So,
$\frac{100}{112} m o l$ of $H_{2} O_{2}$ will give $11.2 \times \frac{100}{112} = 10 L$ of $O_{2}$ at STP

By definition , volume strength of $H_{2} O_{2}$ is $10 V$

Suggest Corrections

Similar questions

8.5 g

280 m L

H_{2} O_{2}

34

1120

H_{2} O_{2}

H_{2} O_{2}

10

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program