Question

❤❤Con.HCI is 38 % HCl by mass.What is the molarity of this solution if d = 1.19 g cm^-3)? What volume of Conc.HCI is required to make 1.00 L of 0.1 0 M HCI?❤❤

Answer 1

Given:-

Mass of HCl = 38 gm

Density = 1.19 g/cm³

Mass of solution = 100 g

Find:-

Molarity of the solution

Solution:-

a)

Molarity is defined as number of moles of solute present in 1 litre of solution.

$\sf{Molarity \:=\:\dfrac{Moles\:of\:HCl}{Volume \:of\:HCl\:in\:litres}}$

⇒ $\sf{\dfrac{w}{m}\:\times\:\dfrac{1000}{Volume}}$

Here..

• w = given mass
• m = molar mass

Substitute the known values in above formula

⇒ $\sf{\dfrac{38}{36.5}\:\times\:\dfrac{1000}{volume}}$ ___ (eq 1)

$\sf{Density \:=\:\dfrac{Mass}{Volume}}$

$\sf{Volume \:=\:\dfrac{Mass}{Density}}$

⇒ $\sf{Volume \:=\:\dfrac{100}{1.19}}$

Substitute value of volume in (eq 1)

⇒ $\sf{\dfrac{38}{36.5}\:\times\:\dfrac{1000}{ \frac{100}{1.19} }}$

⇒ $\sf{\dfrac{38}{36.5}\:\times\:\dfrac{1000 \: \times \: 1.19}{100}}$

⇒ $\sf{1.0411\:\times\:11.9}$

⇒ $\sf{12.38\:M/L (mol\:per\:litre)}$

_______________________________

b)

Given:-

M1 = 12.4 M/L

M2 = 0.10 M/L

V2 = 1.00 lit.

Find:-

Volume of concentrated HCl.

Solution:-

$\sf{M_1\:\times\:V_1\:=\:M_2\:\times\:V_2}$

We have..

• M1 = 12.4 M/L (12.38)
• M2 = 0.1 M/L
• V2 = 1 lit. = 1000 ml

Substitute the given values in above formula

⇒ $\sf{12.4\:\times\:V_1\:=\:0.1\:\times\:1000}$

⇒ $\sf{12.4\:\times\:V_1\:=\:100}$

⇒ $\sf{V_1\:=\:8.06\:ml}$