Question

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

Answer 1

$\checkmark$ Given:

3.65 g of HCI in 500 ml of a solution

$\checkmark$ To Find:

Molarity of the HCl solution

$\checkmark$ Solution:

We know that,

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

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

• V = 500 ml

We know that,

Molar Mass of HCl is 36.5 g

Hence,

• GMW = 36.5 g

By Substituting the values,

We get,

$\blue{\rm \implies M=\dfrac{3.65}{36.5} \times \dfrac{1000}{500}}$

$\green{\rm \implies M=0.1 \times 2}$

$\orange{\rm \implies M=0.2}$

Therefore,

$\pink{\rm \checkmark \ Molarity=0.2 \ M}$

Answer 2

Dear student,

Given :-

• Given mass, w = 3.65 g
• volume of HCl, v(mL) = 500

To find :-

• molarity of HCl solution

Solution :-

We know by the mole concept, that for molarity;

            n' = w/ mw / v(L)

where, n' = molarity

           w = given weight

         mw = total weight

           v = volume ( in L)

According to the question, putting the given values in the above formula, we get;

       n' = 3.65 / mw of HCl × 1000/ v (ml)

       n' = 3.65 / 35.5 + 1 × 1000/500

       n' = 3.65/ 36.5 × 2

       n' = 1/10 × 2

       n' = 1/5

       n' = 0.2

Hence, the molarity of HCl solution is 0.2.