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0.70 g of mixture (NH4)2SO4 was boiled w...

0.70 g of mixture `(NH_4)_2SO_4` was boiled with 100 mL of 0.2 N NaOH solution till all the `NH_3(g)` evolved and get dissolved in solution itself.The remaining solution was diluted to 250 mL.25 mL of this solution was neutralized using 10 mL of a 0.1 N `H_2SO_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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