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Question
why we are taking 85 ml of concentrated HCl to prepare 1N 
Solution
 Question Submitted By :: Analytical-Chemistry
I also faced this Question!!     Answer Posted By  
 
Answer
# 1
According to this formula

Con(mL) =(M.Wt of Hcl * 100)/ (Specific Gravity of Hcl*
Purity of Hcl)

= 36.5 * 100 / 1.18 * 36 = 85.9mL
 
Is This Answer Correct ?    95 Yes 16 No
K.srinivas
 
Answer
# 2
we should consider the density of the solution :

density= M/V
V=M/D=36.5/1.18=30.93 if purity is 100% but in the HCL ,
purity is 36% only , so

Required volume to prepare 1N solution =30.93*100/36=85ml
 
Is This Answer Correct ?    48 Yes 11 No
Kizar Ahamed
 
 
 
Answer
# 3
Normality of Con. HCl =(Purity of Hcl * 10 * specific
gravity of HCl)/ Eq. wt. of HCl

Where 10 is multiplied to convert in normal soln.(1000 mL)

= (36 * 10 * 1.18)/36.5 = 11.64 N.(N1)

Now,N1V1 = N2V2

v1 = (1 * 1000)/11.64
= 85.9 mL
 
Is This Answer Correct ?    39 Yes 4 No
Alpesh Patel
 
Answer
# 4
FIRST WE ARE FIND OUT LAB MORMALITY=(10 X ASSAY X Wt/ml)/M.W

=(10 X 1.18 X 36)/36.5 = 11.64. THEN M1V1=N2V2 SO,V1=(1X1000)/11.64 = 85.91
 
Is This Answer Correct ?    7 Yes 1 No
Ponnam
 

 
 
 
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