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Grade 12Inorganic Chemistry

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%

Profile image of Somya
7 Years agoGrade 12
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1 Answer

Profile image of Kunal Arora
7 Years ago
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)

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