The amount of a caustic soda required for complete neutralisation of $100 m L 0.1 N H C l$ is

4.0 g

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

0.04 g

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

0.4 g

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

2.0 g

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

Open in App

Solution

The correct option is C $0.4 g$
$W = \frac{N o r m a l i t y \times G E W \times v o l u m e}{1000} = \frac{0.1 \times 40 \times 100}{1000} = 0.4$
Hence option C is correct.

Suggest Corrections

Similar questions

What is the amount of sulphuric acid require to neutralise completely 8g of caustic soda? Sir,can you provide me more such questions with answers.

2.0 g

= 600

100 m L

\frac{M}{6} N a O H

4.0 g

(m o l m a s s 40)

How many grams of caustic potash required to completely neutralise 12.6 gm $H N O_{3}$

N a O H

H_{2} S O_{4}

Join BYJU'S Learning Program