Question

Calculate the molarity of a HCI solution, which contains 3.65 g of HCI in 500 ml of a solution​

Answer 1

$\huge\sf\underline\purple{Given:-}$

• 3.65 g of HCI in 500 ml of a solution

$\huge\sf\underline\orange{To\:Find:-}$

• Molarity of the HCl solution

$\huge\sf\underline\green{Solution:-}$

➱We know that,

$\orange{\bigstar \ \boxed{\rm M=\dfrac{W}{GMW} \times \dfrac{1000}{V(ml)}}}$

Here,

• M = Molarity
• W = Weight
• GMW = Gram Molecular Weight
• V = Volume of solution

➱According to the Question,

We are asked to find the Molarity of the HCl solution

Given that,

3.65 g of HCI in 500 ml of a solution

Hence,

• W = 3.65 g

• W = 3.65 gV = 500 ml

We know that,

Molar Mass of HCl is 36.5 g

Hence,

• GMW = 36.5 g

➱By Substituting the values,

We get,

$M= 3.65/36.5 × 1000/500$

⟹M=0.1×2

⟹M=0.2

Therefore,

✓ Molarity=0.2 M

Answer 2

Given mass of HCl = 3.65 g

Molar mass of HCl = 36.5 g/mol

Basicity of HCl = 1

Now , equivalent of HCl =Molar mass of HCl÷Basicity

= 36.5 ÷ 1 = 36.5 g /equiv

We have to find out the total number of equivalents of HCl.

So, Mass of HCl ÷ Equivalent mass of HCl

= 3.65 ÷ 36.5 = 0.1 equivalent

Given volume of solution = 500 ml = 0.5 l

Normality = Equivalents of HCl ÷ Volume of solution(l)

= 0.1 ÷ 0.5 = 0.2 N