Question

The molarity and molality of solution of h2so4 (specific gravity is 1.3) containing 100 g/litre of h2so4 is:

Answer 1

Hello Dear.

☆ Here is the answer ☆

Given
Specific gravity = 1.3

Since,
Specific Gravity = Density of the solution/Density if the Water.

Now, Density of Water = 1 g/ml.

Density of Water = 1.3 × 1 g/ml.
= 1.3 g/ml.

Density of Sulphuric Acid = 100 g/liter.
It means that 100 g of suphuric acid is present in 1 liter of solution.

Since,
No. Of moles of sulphuric acid = Mass/Molar Mass.
= 100/98
= 1.02 moles.

For Calculating Molarity,

Using the Formula,

Molarity = No. Of moles of solute/Volume of the solution in liter.
= 1.02/1
= 1.02 M.

For Calculating Molality,

Volume of the Solution = 1 liter.
= 1000 ml.
Density of the Solution = 1.3 g/ml.

Therefore,
Mass of the solution = 1.3 × 1000
= 1300 g.

Now, Mass of the Solvent = Mass of the Solution - Mass of the Solute.

Mass of the Solvent = 1300- 100
= 1200 g.
= 1.2 kg.

Using the Formula,

Molality = No. Of moles of Solute/Mass of the Solvent in kg.
= 1.02/1.2
= 0.84 m.

Thus, Molality of the solution is 0.84 m.


☆ ■ Hope it helps. ■ ☆

Answer 2

Specific gravity is the ratio of density of object to the density of water.
we know, density of water = 1g/mL
so, 1.3 = density of solution/density of water
density of solution = 1.3 × 1g/ml = 1.3 g/mL

Given, 100g H₂SO₄ in 1000 ml of solution .
molarity = number of mole of solute/volume of solution in L
number of mole of solute = given weight/molecular weight
= 100/98 = 1.02
Now, molarity = 1.02/1 = 1.02 M [because volume solⁿ = 1000ml= 1L]
∴molarity = 1.02 M.

Now, use formula ,
1/m = d/M - Ms/1000
here, m is molality , M is Molarity , d is density of solution and Ms is molar mass of solute.
∴ 1/m = 1.3/1.02 - 98/1000
= 1.27 - 0.098
= 1.172
so, m = 1/1.172 m = 0.8532 m
Hence, molality = 0.8532 m