MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Equivalent Mass

A solution co...

Question

A solution contains $N a_{2} C O_{3} a n d N a H C O_{3}$ .10 ml of the solution requires 2.5 ml of 0.1M $H_{2} S O_{4}$ for neutralisation using phenolphthalein as an indicator. Methyl orange is added when further 2.5 ml of 0.2 M $H_{2} S O_{4}$ is required. The amount of $N a H C O_{3}$ in one litre of the solution is

A

2.1g

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

B

8.4 g

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

C

5.3 g

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

D

4.2 g

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

Open in App

Solution

The correct option is D 4.2 g
$N o . m e q N a_{2} C O_{3} \equiv N o . m e q N a H C O_{3} \equiv 2.5 \times 0.1 \times 2 = 0.5 m e q N o . o f m e q N a H C O_{3} + N o . o f M e q N a H C O_{3} = 2.5 \times 0.2 \times 2 = 1.0 m e q (f o r m e d f r o m N a_{2} C O - 3) 0.5 + N o . o f m e q N a H C O_{3} = 1.0 N o . o f m e q N a H C O_{3} = 0.5 m e q \frac{W}{84} \times 1000 = 0.5, W = \frac{0.5 \times 84}{1000} 10 m l o f s o l u t i o n h a s \frac{0.5 \times 84}{1000} g 100 m l - \frac{0.5 \times 84}{1000} \times \frac{1000}{10} = 4.2 g / l i t$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A solution contains

N a_{2} C O_{3}

N a H C O_{3}

10 m L

of the solution required

2.5 m L

0.1 M H_{2} S O_{4}

for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further

2.5 m L

0.2 M H_{2} S O_{4}

was required. Calculate the amount of

N a_{2} C O_{3}

N a H C O_{3}

in one litre of the solution.

50 m L

of a solution of

N a_{2} C O_{3}

N a H C O_{3}

17.5 m L

0.1 M H_{2} S O_{4}

for neutralisation using phenolphthalein as indicator. Methyl orange is then added when further

25 m L

0.2 M H_{2} S O_{4}

was required. Calculate the weight of

N a_{2} C O_{3}

N a H C O_{3}

respectively in the solution.

50 m L

of a solution of

N a_{2} C O_{3}

N a H C O_{3}

17.5 m L

0.1 M H_{2} S O_{4}

for neutralisation using phenolphthalein as indicator. Methyl orange is then added when further

25 m L

0.2 M H_{2} S O_{4}

was required. Calculate the weight of

N a_{2} C O_{3}

N a H C O_{3}

respectively in the solution.

Q. A solution contain

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

N a H C O_{3}

. 10 ml of this solution required 2.5 ml of 0.1 M

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

for neutralization using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M

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

was required. The amount of

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

N a H C O_{3}

in 1 litre of the solution is ?

Q. A solution contained

N a_{2} C O_{3}

N a H C O_{3}

15 m l

of the solution required

5 m l

\frac{N}{10} H C l

for neutralization using phenolphthalein as an indicator. Addition of methyl orange required a further

15 m l

of the acid for neutralization. The amount of

N a_{2} C O_{3}

present in the solution is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Reactions in Solutions

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Equivalent Mass

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon