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25 mL 0.1 N H(2)SO(4) neutralized with 2...

25 mL 0.1 N `H_(2)SO_(4)` neutralized with 20 mL `xN Na_(2)CO_(3)`. What will be the g/liter of `Na_(2)CO_(3)` ?

A

8.48 g

B

4.24 g

C

6.625 g

D

13.25 g

Text Solution

Verified by Experts

The correct Answer is:
B, C
`N_(1)V_(1) =N_(2)V_(2)`
`(0.1)(25) =(x)(20)`
`:.x=(0.1xx25)/(20)`
`:.N_(2)=0.125`
`:.0.125 N Na_(2)CO_(3) = (0.125)/(2)`
`= 0.0625 M Na_(2)CO_(3)` solution
`= 0.0625xx10^(6) g//L`
`=6.625 g`
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