The molality of `1 L` solution with `x% H_(2)SO_(4)` is equal to 9. The weight of the solvent present in the solution is `910 g`. The value of `x` is:

The molality of 1 L solution with x% H_(2)SO_(4) is equal to 9. The weight of the solvent present in the solution is 910 g . The value of x in g per 100 mL is

The molality of a 1 L solution with x % H_2SO_4 is 9. The weight of solvent present in the solution is 910 grams. The value of x is :

The molality of X% H_(2) SO_(4) solution is equal to 9. The weight of the solvent present in the solution is 910g.The value of 'X' is : 90, 80.3, 40.13, 9

The molality of an H_(2)SO_(4) solution is 9. The weight of the solute in 1 kg H_(2)SO_(4) solution is:

Calculate the molality of 1 L solution of 93% H_(2)SO_(4) (Weight/volume) The density of the solution is 1.84 g .

Calculate the molality of 1 L solution of 93% H_(2)SO_(4) (Weight/volume) The density of the solution is 1.84 g .