Question

14.2g of Na2SO4 is added to form 800 ml of solution.what is the molarity of solution?​

Answer 1

Solution:

Mass of Na₂SO₄ = 14.2 g

Mass of one mole of Na₂SO₄ = 142.04 g/mol ≈ 142 g/mol

Number of moles of Na₂SO₄ = $\frac{Mass \ of \ substance(Na_{2} SO_{4} )}{Mass \ of \ one \ mole \ of \ substance(Na_{2} SO_{4} ) }$ = $\frac{14.2}{142}$ = $0.1 \ moles$

Volume of solution = 800 ml = 800/1000 = 0.8 litre

Molarity of Solution(M) = $\frac{Moles \ of \ Solute(Na_{2}SO_{4} )}{Volume \ of \ Solution \ in \ litre}$ = $\frac{0.1}{0.8} = 0.125$

Molarity of solution(M) as we can see in the formula is the number of moles of solute which gets dissolved in one liter of solution.

Answer 2

Answer- The above question is from the chapter 'Some Basic Concepts of Chemistry'.

Molarity: Number of moles of solute present in 1 litre of solution is called molarity.

Unit of Molarity = Molar (M)

•When volume is given in litres,

Molarity = Number of moles of solute ÷ Volume of solution in litres

•When volume is given in millilitres,

Molarity = (Number of moles of solute × 1000) ÷ Volume of solution in ml

•When mass % and density is given,

Molarity = (Mass % × Density × 10) ÷ Molar mass of Solute

Given question: 14.2g of Na₂SO₄ is added to form 800 ml of solution. What is the molarity of solution?​

(Atomic masses of Na = 23 u, S = 32 u and O = 16 u.)

Answer: Given mass of Na₂SO₄ = 14.2 g

Gram molecular mass of Na₂SO₄ = 2[Na] + 1[S] + 4[O] = 46 + 32 + 64

                                                          = 142 g

Number of moles of Na₂SO₄ = Given mass/Gram atomic/molecular mass

                                                = 14.2/142 = 1/10 = 0.1 moles

Volume = 800 ml

Molarity = (Number of moles of solute × 1000) ÷ Volume of solution in ml

∴ Molarity = (0.1 × 1000)/800 = 1/8 = 0.125 Molar or 0.125 M

∴ If 14.2 g Na₂SO₄ is dissolved in 800 ml of solution, its molarity is 0.125 M.