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

Bleaching Powder

The solution ...

Question

The solution having lowest molar concentration is :

1 N H C l

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

0.4 N H_{2} {S O}_{4}

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

0.1 N {N a}_{2} {C O}_{3}

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

1 N N a O H

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

Open in App

Solution

The correct option is C $0.1 N {N a}_{2} {C O}_{3}$
$M o l a r c o n c e n t r a t i o n = \frac{N o r m a l c o n c e n t r a t i o n}{e q u i v a l e n t f a c t o r}$
$F o r 1.0 N H C l, M_{1} = \frac{1}{1} = 1 M$

$F o r 0 \cdot 4 N H_{2} S O_{4}, M_{2} = \frac{0 \cdot 4}{2} = 0 \cdot 2 M$

$F o r 0 \cdot 1 N N a_{2} C O_{3}, M_{3} = \frac{0 \cdot 1}{2} = 0 \cdot 05 M$

$F o r 1 N N a O H, M_{4} = \frac{1}{1} = 1 M$
Hence, $0.1 N$ $N a_{2} C O_{3}$ has the lowest concentration.

Therefore, option C is correct.

Suggest Corrections

Similar questions

100

1 N H C l

50

1 N N a O H

10^{- 3} N

Heat liberated when 100 ml of 1 N NaOH is neutralized by 300 ml of 1N HCl is

100 m l

{N a}_{2} {C O}_{3}

5.3 g

{N a}_{2} {C O}_{3}

200 m l

H_{2} {S O}_{4}

H_{2} {S O}_{4}

l o g 5 = 0.6990

H^{+}

5 \times 10^{- 3} M H_{2} {C O}_{3}

10

1.25 g

{N a}_{2} {C O}_{3}

{N a}_{2} {S O}_{4}

250 m l

25 m l

20 m l

0.1 N

H_{2} {S O}_{4}

{N a}_{2} {C O}_{3}

Join BYJU'S Learning Program