$N a C l O$ solution reacts with $H_{2} S_{4} O$ . Solution of $N a C l O$ used in the reaction contained 15g of NaClO per litre. The normality of the solution would be:

0.8

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

0.6

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

0.2

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

0.4

(approx)

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

Open in App

Solution

The correct option is D $0.4$ (approx)
$N a c l O = 74 \cdot 5$
$m o l a r i t y = \frac{15}{74 \cdot 5} \times \frac{1000}{1000}$
$= 0 \cdot 201$
$N o r m a l i t y = m o l a r i t y \times 2$
$= 0 \cdot 402$

Suggest Corrections

Similar questions

N a C l O

H_{2} S O_{3}

N a C l O + H_{2} S O_{3} \to N a C l + H_{2} S O_{4}

N a C

15 g

N a C l O

The product(s) when dichlorine hexoxide reacts with $N a O H$ is(are):

{N H}_{4} {C l O}_{4}

{N a C l O}_{4}

1.0 k g

N a C l + 4 H_{2} O ⟶ {N a C l O}_{4} + {4 H}_{2}

Electrolysis of fuel NaCl will give

OPTIONS

1) Na 3) NaClO

2) NaCl 4) NaClO₃

$C l_{2} + N a O H \to N a C l + A - - + H_{2} O A$ is

(hot & conc)

Join BYJU'S Learning Program