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Concentrated HCl solution is 37.0% HCl and has a density of 1.19 g/mL. A dilute solution of HCl is prepared by diluting 4.50mL of this concentrated HCl solution to 100mL with water. Then 10mL of this dilute HCl solution reacts with an AgNO3 solution. Calculate the volume of 0.108M AgNO3 solution required to precipitate all the chloride as AgCl(s).


Answer
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Hint: The formula to find molarity and molarity of dilute solution obtained from concentrated solution are as below.
M=Weight of solute×1000Molecular weight of solute×Volume of the solute(mL)
M1V1=M2V2

Complete Step-by-Step Solution:
We will first calculate the molarity of concentrated HCl solution. From that, we will find the molarity of dilute HCl solution and finally from it, we will find the volume of AgNO3 solution required to complete precipitation.
- We are given that the concentrated HCl solution is 37%. So, we can say that if the solution is of 100 g, then the weight of dissolved HCl will be 37g.
- Density of the solution is given 1.19 g/mL. Now, we will use the formula of the density given below.
Density = WeightVolume
So, 1.19=100Volume
Thus, we can write that Volume = 1001.19=84.03mL
We know that Molecular weight of HCl = Atomic weight of H + Atomic weight of Cl
So, Molecular weight of HCl = 1 + 35.5 = 36.5 g/mol
Now, we will find the molarity of the solution.
M=Weight of solute×1000Molecular weight of solute×Volume of the solute(mL)
So,
M=37×100036.5×84.03=12.06M
Now, we will find the molarity of diluted solution of HCl using the equation given below.
M1V1=M2V2
where M1 = Molarity of con. HCl solution = 12.06M
V1 = volume of cone. HCl solution taken to make dilute solution = 4.50 mL
M2 =Molarity of diluted solution
V2 =Volume of diluted solution of HCl = 100mL
So, we can write that
12.06×4.50=M2×100
So,
M2=12.06×4.50100=0.5427M
Now, we will take a look at the reaction between HCl and AgNO3.
HCl+AgNO3AgCl+HNO3
 - Now, we are given that 10mL of this 0.5427M solution is allowed to react with 0.108M AgNO3 solution.
- From the definition of molarity, we can say that 1000mL of 0.5427M solution will contain 0.5427 moles of HCl. So, 10mL of 0.5427M solution will contain 0.005427 moles.
- Now, we will require 0.005427 moles of AgNO3 to have complete precipitation.
- From the definition of molarity, we can say that 0.108 moles of AgNO3 will be obtained from 1000mL of 0.108M AgNO3 solution. So, 0.005427 moles of AgNO3 can be obtained from 0.005427×10000.108=50.25mL

Thus, we found that we will require 50.25mL of 0.108M AgNO3 solution to ensure complete precipitation.

Note: Remember that whenever we make a solution from another solution of known concentration, then we can use the formula M1V1=M2V2 to find the volume or molarity whichever is unknown. If we are given normality, then we can use normality of both the solutions in the same formula.
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