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5gm impure calcium hydroxide is dissolved into 800 ml of water. In 100ml of this solution 20ml of decinormal HCl is added.The acidic solution formed is neutralized by 50ml of N/50 NaOH solution.Percentage purity of Ca(OH)2 is,
1) 44%
2) 5.92%
3) 14.34%
4) 7.5%

Somya , 6 Years ago
Grade 12
anser 1 Answers
Kunal Arora
we have given that 5g of Ca(OH)2  is added in 800 ml of H2O.
so normality of Calcium hydroxide is .
N= (W ×1000)÷ (Eq × volume)
N=(5×1000)÷(37×800)    ( eq = 74/2)
N=0.1689N
 now on mixing 20 ml of 0.1 N HCl and 50 ml of N/50 NaOH in it we can use equatiom of mixing i.e.
[N1V1(HCl)- N2V2(NaOH)-N3V3(Ca(OH)2)]/V1+V2+V3=N
 
for neutral solution N=10*-7
as pH = 7 
 
 also let N of Ca(OH)2 is N3 .
so putting values in equation.
[(0.1×20/1000)-(1/50× 50×1000)-(N3×100÷1000)]/170=10*-7
  
on solving this equation we get N3=9.83×10*-3
 
so percentage purity is
   N(observed)/N(calculated)×100
=(9.83×10*-3/168.9×10*-3)×100
=5.92  
 
 so in my views answer is (2)
Last Activity: 6 Years ago
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