1.7 g of ammonium salt was heated with excess of NaOH. The ammonia released in the process neutralises 100 c.c. solution of N/5 H2SO. What is the percentage of ammonia in the salt?
1) 20%
2) 35%
3) 1.7%
4) 40%
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Answer:
Explanation:
=> For, 1L (1000 ml ) solution
Normality, N1 =[(Wb/Mb)∗1000/ V]
=(1.7/17.03)×1000/1000
=0.1N
=> suppose the percentage of purity is x
N1= initial Normality = 0.1N
N2= Normality of the new solution = 1/5 N
V1= initial volume = 1000
V2= volume of the new solution = 100
the factor of purity = x/100
=> Let N1V1=N2V2
∴ (x/100)[0.1∗1000]=1/5∗100
x = 1 * 100 * 100 / 5 * 0.1 *1000 = 20
x = 20%
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