You visited us 1 times! Enjoying our articles? Unlock Full Access!

Hydrogen Peroxide

Excess of K...

Question

Excess of $K I$ and dil. $H_{2} S O_{4}$ , were mixed in $50$ ml $H_{2} O_{2}$ . The $I_{2}$ liberated required $20$ ml of $0.1$ N $N a_{2} S_{2} O_{3}$ . Find out the strength of $H_{2} O_{2}$ in $g / l i t^{- 1}$ ?

0.17 g / l i t

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

0.34 g / l i t

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

0.68 g / l i t

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

1.36 g / l i t

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

Open in App

Solution

The correct option is C $0.68 g / l i t$
$I_{2} + 2 N a_{2} S_{2} O_{3} - > N a_{2} S_{4} O_{6} + 2 N a I$
20ml of 0.1M $N a_{2} S_{2} O_{3}$ will react with $\frac{(20) (0.1)}{(1000) (2)}$ moles of $I_{2}$ which is 0.001 moles or 0.002 equivalents. Thus there are 0.002 equivalents in 50ml of $H_{2} O_{2}$ . Hence, its strength is 0.002 x 17/0.05=0.68

Suggest Corrections

Similar questions

20

H_{2} O_{2}

H_{2} S O_{4}

30

K M n O_{4}

H_{2} O_{2}

K I

d i l . H_{2} S O_{4}

50 m L H_{2} O_{2}

I_{2}

20 m L

0.1 N N a_{2} S_{2} O_{3}

H_{2} O_{2}

g p e r l i t r e

50

H_{2} O_{2}

K I

H_{2} S O_{4}

20

0.1

N N a_{2} S_{2} O_{3}

H_{2} O_{2}

25

H_{2} O_{2}

K I

20

0.1 N

N a_{2} S_{2} O_{3}

H_{2} O_{2}

H_{2} O_{2}

N a_{2} S_{2} O_{3}

H_{2} O_{2}

Join BYJU'S Learning Program