Molality

Let us learn about "molality"

An alternative unit of concentration to molarity is molality. The molality of a solute is the number of moles of that solute divided by the weight of the solvent in kilograms. For water solutions, 1 kg of water has a volume close to that of 1 liter, so molality and molarity are similar in dilute aqueous solutions.

The 'n' of a solute present per litre of the solution is called as Molarity, M of the solution .

Solved Problems on Molarity

1) Calculate the weight of Na₂CO₃ present in 100 ml of 0.1 M Na₂CO₃ solution ?

Solution : Volume of the solution = 100 ml = V ml

Weight of Na₂CO₃ dissolved = 'G' g (say)

Gram molecular weight of Na₂CO₃ = { (23*2) + 12 + (16*3) } = 106 g = GMW

Molarity of the given solution = 0.1 M = M

Substituting the values in the formula

M = G/GMW * 1000/V(ml)

0.1 = G/106* 1000/100

G = 0.1*106*100/1000 = 1.06 g

Weight of Na₂CO₃ = 1.06 g .

in our next blog we shall learn about exothermic reaction

I hope the above explanation was useful.Keep reading and leave your comments.