Commercially available concentrated Hcl contains 38% Hcl by mass. (1) what is the molarity of the solution ( density of solution =1.19 g mL -1)

(2) What volume of concentrated Hcl is required is required to make 1.0 L of an 0.10 M Hcl.

A 38% by mass solution of HCl means that 38 g of HCl is present in 62g of water, thereby making the total mass of solution equal to 100g. The molar mass of HCl is 36.5g. Therefore, the number of moles of HCl that are present in 38g of HCl are
= mass of HCl / molar mass of HCl
= 38 / 36.5
= 1.04
We known, Molarity = Number of moles/ Volume of solution (in L)
As Volume of solutions is not known therefore, we will take help of density

Density = mass / volume
Mass of solution = Mass of solute + Mass of solvent 
= 100 g 
Volume = 100/ 1.1 
= 90.91 mL

​On substituting the above values in the formula of molarity, we will get 
M = 1.04 ×1000/ 90.91
= 11.44 M

​2. 
The molarity of 38 % solution is 11.44 M
As we are supposed to make it 0.10 M in 1 L, therefore we can use the formula 
M1V1 = M2V2  ...(ii)
Where 1 stands for commercially available HCl and 2 stands for final solution.

M1 = 11.44 M 
M2 = 0.10 M 
V2 = 1 L 
on substituing the above value in equation (ii), we will get 
V1 = 0.10 × 1  /  11.44 
= 0.0087 L 
or 8.74 mL of HCl would be required. 

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1) Concentrated HCl 38% by mass i.e 100gm of HCL soclution contains 38gm of HCl moles of HCl = given mass/molar mass = 38/36.5 Now density of solution = 1.19 g mL-1Since solution is 100gm. Density = Mass/volume = Volume= mass/density = 100/1.19 ml Molarity = Moles of Solute/volume of Solution in L M = (38/36.5)*(1000*1.19/100) = 12.38 M

2) M = Given Mass/Molecular mass* volume of solution in L 0.10M = x/36.5*1 = 0.10M * 36.5 = x = x = 3.65 gm Moles of HCl = 3.65/36.5 = 0.1 mole at STP 1mole = 22.4 L, so 0.1 mole = 2.24 L

Ans1 = 12.38M

Ans2 = 2.24 L

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