calculate the molarity of 23% of h2so4 when density is 1.4 g/ml
calculate the molarity of 23% of h2so4 when density is 1.4 g/ml
A 23% solution of H2SO4 means that 23g of sulphuric acid is present in 77g of water, thereby making the total mass of the solution equal to 100g. Thus
mass of sulphuric acid present = 23g
The molarity of a solution is given by the formula
molarity = number of moles of solute / Volume of solution (in litres)
= (mass of solute / molar mass of solute) X [1 / volume of solution (in litres)]
The volume of sulphuric acid solution will be calculated from its density by using the relation
Density = mass / Volume
We are given that the density of sulphuric acid solution is 1.4 g/ml.
So volume = mass / density = 100 / 1.4 = 71.428 ml
Thus molarity of 23% sulphuric acid solution will be
= (23 / 98) X (1000 / 71.428)
= 3.286 M
A 23% solution of H2SO4 means that 23g of sulphuric acid is present in 77g of water, thereby making the total mass of the solution equal to 100g. Thus
mass of sulphuric acid present = 23g
The molarity of a solution is given by the formula
molarity = number of moles of solute / Volume of solution (in litres)
= (mass of solute / molar mass of solute) X [1 / volume of solution (in litres)]
The volume of sulphuric acid solution will be calculated from its density by using the relation
Density = mass / Volume
We are given that the density of sulphuric acid solution is 1.4 g/ml.
So volume = mass / density = 100 / 1.4 = 71.428 ml
Thus molarity of 23% sulphuric acid solution will be
= (23 / 98) X (1000 / 71.428)
= 3.286 M
