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Submit your detailed Question here...29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Submit your detailed Question here...29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.
The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Grade:11

2 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
8 years ago
29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is as follows
stock solution of HCl=29.2%(w/w)
molarity of stock solution of HCl=29.2*1000*1.25/36.5*100
if volume of stock solution required is Vml
then
v*29.2/36.5*1000/80=200*.4
v=8ml
Urvi
13 Points
3 years ago
M1= 29.2*1.25*10/36 .5
M1=10 
M2=0.4
V2=200
M1V1=M2V2
10*V1=0.4*200
V1=0.4*200/10
V1=8
So the answer is 8 ml 
And this is solved by the formula 
Molarity=%(w/w) *density of solution*10/molar mass of solute 
 
 
 

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