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0.7g of (NH(4))(2)SO(4) sample was boile...

0.7g of `(NH_(4))_(2)SO_(4)` sample was boiled with 100mL of 0.2 N NaOH solution was diluted to 250 ml. 25mL of this solution was neutralised using 10mL of a 0.1 N `H_(2)SO_(4)` solution. The percentage purity of the `(NH_(4))_(2)SO_(4)`sample is :

A

94.3

B

50.8

C

47.4

D

79.8

Text Solution

Verified by Experts

The correct Answer is:
A
m.eq of `(NH_(4))_(2)SO_(4)+` m.eq of `H_(2)SO_(4)`=m.eq of NaOH
`("m.molea"xx2)++(0.1xx10xx(250)/(25))=0.2xx100`
`:. " m.mole of " (NH_(4))_(2)SO_(4)=5`
wt. of `(NH_(4))_(2)SO_(4)=(5)/(1000)xx132=0.66 g`
`:. " % of " (NH_(4))_(2)SO_(4)=(0.66)/(0.7)xx100=94.28% ~~94.3%`
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