Question

300ml of 20% w/w% na2so4 (aq) solution ,find molarity?​

Answer 1

Given :-

◉ 300 ml of 20% of Na₂SO₄ aqueous solution (w/w)

To Find :-

◉ Molarity of the solution

Solution :-

We know,

⇒ Molarity = Number of moles of solutes / Volume of solution (in L)

So, We should have number of moles of Na₂SO₄ and volume of solution (which is 300 mL)

We have, Mass of :-

• Sodium, Na = 23 g
• Sulphur, S = 32 g
• Oxygen, O = 16 g

Now, molar mass of Na₂SO₄

⇒ M = 23×2 + 32×1 + 16×4

⇒ M = 46 + 32 + 64

⇒ M = 142 g

We are given the weight by weight percentage of Na₂SO₄ solution as 20% which means

• 20 g Na₂SO₄ is present in 100 g of solution.

So, we have

⇒ 20 g of Na₂SO₄ ➞ 100 ml of solution

In 1 ml solution, 1/5 g Na₂SO₄ is present.

Hence, In 300 ml Solution, 60 g Na₂SO₄ is present.

Let us find the number of moles of Na₂SO₄,

⇒ Number of moles = given mass / molar mass

⇒ No. of moles = 60 / 142

⇒ No. of moles = 0.42 mole

Further, Volume of solution is 0.3 , so we have

⇒ Molarity = 0.42 / 0.30

⇒ Molarity = 1.4 M

Answer 2

Answer:

Given :-

◉ 300 ml of 20% of Na₂SO₄ aqueous solution (w/w)

To Find :-

◉ Molarity of the solution

Solution :-

We know,

⇒ Molarity = Number of moles of solutes / Volume of solution (in L)

So, We should have number of moles of Na₂SO₄ and volume of solution (which is 300 mL)

We have, Mass of :-

Sodium, Na = 23 g

Sulphur, S = 32 g

Oxygen, O = 16 g

Now, molar mass of Na₂SO₄

⇒ M = 23×2 + 32×1 + 16×4

⇒ M = 46 + 32 + 64

⇒ M = 142 g

We are given the weight by weight percentage of Na₂SO₄ solution as 20% which means

20 g Na₂SO₄ is present in 100 g of solution.

So, we have

⇒ 20 g of Na₂SO₄ ➞ 100 ml of solution

In 1 ml solution, 1/5 g Na₂SO₄ is present.

Hence, In 300 ml Solution, 60 g Na₂SO₄ is present.

Let us find the number of moles of Na₂SO₄,

⇒ Number of moles = given mass / molar mass

⇒ No. of moles = 60 / 142

⇒ No. of moles = 0.42 mole

Further, Volume of solution is 0.3 , so we have

⇒ Molarity = 0.42 / 0.30

⇒ Molarity = 1.4 M