If `250 mL` of a solution contains `24.5 g H_(2)SO_(4)` the molarity and normality respectively are:

If 100 ml of a solution contains 10 g of H_(2)SO_(4) , normality of the solution is

A solution contains 441.0 gof H_2SO_4 in 1 L of solution. If the density of solution is "1.25 g mL"^(-1), the molarity and molality of the solution, respectively, are

A 25mL HCl solution containing 3.65g HCl/L is neutralized by 50mL of NaOH solution. Again, 25mL of an H_(2)SO_(4) solution of unknow strenght. The normally of the H_(2)SO_(4) solution is :

Converting mass percent to molarity: the density of a 24.5 mass % solution of sulphuric acid (H_(2)SO_(4)) in water is 1.176 g mL^(-1) at 25.0^(@)C . What is the molarity of the solution? Stregy: Use mass % to calculate number of moles of solute and use density of solution to calculate volume of solution. Describing a solution as 24.5 mass % H_(2)SO_(4) in H_(2)O means that every 100.0 g of solution contains 24.5 g of H_(2)SO_(4) and 75.5 g of H_(2)O . Since we want to calculate the concentration in molarity, we first need to find the number of moles of H_(2)SO_(4) dissolved in a specific mass of solution. Next we use density as a conversion factor to find the volume of that solution and then calculate molarity by dividing the number of moles of H_(2)SO_(4) by the volume of solution.

Calculate the normality and molarity of H_2SO_4 solution containing 4.9 g of H_2SO_4 per litre of the solution.

The molarity of the solution containing 7.1g of Na_(2)SO_(4) in 100mL of aqueous solution is

5 g sample contain only Na_(2)CO_(3) and Na_(2)SO_(4) . This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M H_(2)SO_(4) . Calcalute the % of Na_(2)SO_(4) in the sample .