Molarity of `H_(2)SO_(4)` is 0.8 and its density is `1.06g//cm^(3)`. What will be its concentration in terms of molality and mole fraction?

Molarity of H_(2)SO_(4) is 0.1 M and its density is 1.06 g cm^(-3) . What will be concentration of the solution in terms of molarity and mole fraction ?

Molarity of H_(2)SO_(4) is 18 M. Its density is 1.8 g//cm^(3) ,hence molality is:

Molarity of H_(2)SO_(4) is 18 M . Its density is 1.8 g//cm^(3) , hence molaity is :

The molarity of H_(2)SO_(4) is 18 M . Its density is 1.8 g mL^(-1) .

Molarity of 15 % H_(2) SO_(4) of density 1.1 g / cm^(3) is _________.

Converting molarity to mole fraction, mass percent and molality: A 0.750 M solution of H_(2)SO_(4) in water has a density of 1.049 g mL^(-1) at 20 ^(@)C . What is the concentration of this solution in (a) mole fraction, (b) mass percent. And (c ) molality? Strategy: To solve this problem, we need to know the mass of solute as well as mass of solvent. There is 0.750 mol H_(2)SO_(4) per liter of solution. Let's pick an arbitrary amount of the solution that will make the calculations easy, say 1.00 L . Consider 1 L (=1000 ml) of solution and calculate its mass, using the density. We can then calculate the mass of H_(2)SO_(4) and find the mass of H_(2)O by difference. Now we can calculate the mole fraction, mass precent and morality.