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Question

2.20 g of an ammonium salt was boiled with 75 mL of NaOH till the emission of ammonia gas ceased. The excess of unused NaOH solution required 70 mL of N/2 sulphuric acid for neutralisation. Calculate the percentage of ammonia in the salt.

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Solution

(NH4)xA+NaOH(Na)xA+NH3+H2O

Volume of NaOH consumed by H2SO4 = y

N×Y=N10×70

Y= 35mL

Out of 75mL , 35mL consumed by H2SO4

Hence, 40 mL consumed by NH3.

Amount of NH3 in N NaOH = Eq. mass of NH3×401000
= 17×401000 = 0.68g.


Percentage of ammonia in salt = 0.6820×100 = 30.9 %


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