The relation between molarity (M) and molality (m) is given by : (p=density of solution (g/mL), `M_1`= molecular mass of solute)

Find the molarity and molality of a 15% solution of H2S04 (density of H_2SO_4 solution = 1.10 g cm^(-3) , Molecular mass of H_2SO_4 = 98 ).

The expression relating molarity (M) of a solution with its molarity (m) is

Molarity is related to molality (m) , density of the solution (d) and molar mass of the solute ( M^(@)) by which of following relation.

Calculate the molarity (M) and molality (m) of 16% aqueous methanol (CH_(3) OH) solution by volume. Density of solution = 0.9 g mL^(-1) .

The molarity of NaNO_(3) solution is 1 M. The density of solution is 1.25 g/ml. Calculate molality of solution. (Molar-Mass of NaNO_(3)=85 g ).

A solution of glucose ("molar mass"= 180g mol^(-1)) in water is labelled as 10% (by mass). What would be the molarity and molality of the solution? Given that the density of the solution is 1.2g mL^(-1) .

The molarity and molality of an aqueous solution of H_(2)SO_(4) "as" M_(1) and M_(2) respectively. If the molecular mass of H_(2)SO_(4) is M, density of the solution is

The expression relating mole fraction of solute (chi_(2)) and molarity (M) of the solution is: (where d is the density of the solution in g L^(-1) and Mw_(1) and Mw_(2) are the molar masses of solvent and solute, respectively

Assertion :- The molarity of a certain solution is always greater than molority. If density of solution is nearly 1gm//ml . Reason :- m=(1000M)/(1000M-MxxM_(solute))

If molarity = molality and density of the solution is 2gm/ml then W_("solute"):W_("solvent") would be