Question

the molarity of sulphuric acid is 2.32 mol/dm3 .if the density is the solution is 1.14 g/cm3 . the molality of solutions is​

Answer 1

Answer:

Explanation:

MOLARITY= 2.32 MOL/DM3= 2.32 MOLES/LITRE= 2.32 MOLAR.

SO WE HAVE 2.32 MOLES OF SOLUTE IN 1L(1000ML) OF SOLUTION.

MASS OF SULPHURIC ACID= 2.32 X 98 GRAM= 227.36 GRAM

NOW DENSITY= 1.14 g/cm3.

MASS OF SOLUTION= DENSITY X VOLUME= 1.14 X 1000= 1140 GRAM.

MASS OF SOLVENT= 1140-227.36= 912.64 GRAM. = 912.64/1000 KG.

MOLALITY = 2.32/ 912.64/1000= 2.54 MOLAL.

Answer 2

Answer:

The molality of given solution is equal to 2.54molal.

Explanation:

Given : The molarity of sulphuric acid $= 2.32moldm^{-3}$

It means 2.32 moles of $H_{2}SO_{4}$ are dissolved in 1000ml of solution.

Number of moles of H₂SO₄ $= 2.32moles$

Weight of  H₂SO₄ = (moles).(molar mass)

                             $=(2.32)(98)\\=227.36g$

Given, density of the solution $=1.14gcm^{-3}=1.14gml^{-1$

Weight of solution = Density× volume $=(1.14)(1000)=1140g$

Weight of water (solvent) $=1140-227.36=912.64g=0.91264kg$

Molality of solution = (moles of H₂SO₄)/(weight of water in Kg)

Molality,  $m=\frac{2.32mol}{0.91264kg}$

Molality of solution $=2.54molkg^{-1}$

Therefore, the molality of solution is 2.54m.