`1.25g` of a solid dibasic acid is completely neutralised by `25mL` of `0.25` molar Ba `(OH)_2` solution. Molecular mass of the acid is :

0.45g of an acid of mol. Mass 90 was neutralised by 20mL of 0.5 N caustic potash (KOH) . The basicity of acid is :

100cm^3 of a solution of an acid (Molar mass =98 ) containing 29.4g of the acid per litre were completely neutralized by 90.0 cm^3 of aq. NaOH containing 20g of NaOH per 500cm^3 . The basicity of the acid is :

25mL of 2N HCI, 50 mL of 4N HNO_3 and x mL of 2MH_2 SO_4 are mixed together and the total volume is made up to 1L after dilution. 50mL of this acid mixture completely reacted with 25mL of a 1NNa_2 CO_3 solution. The value of x is :

Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1

0.80g is impure (NH_4)SO_4) was boiled with 100mL of 0.2N NaOH solution till all the NH_3 (g) evolved. The remaining solution was diluted to 250mL. 25mL of this solution was neutralized using 5mL of 0.2 NH_2 SO_4 solution. The percentage purity of the (NH_4)_2SO_4 sample is :

1.26 g of hydrated oxalic acid was dissolved in water to prepare 250ml of solution. calculate molarity of the solution.

The coagulation of 100 mL of a colloidal solution of gold is completely prevented by the addition of 0.25 g of starch to it before adding 1 mL of 10% NaCl solution . Calculate the gold number of starch.

2.5 litre of 1 M NaOH solution are mixed and another 3 litre of 0.5 M NaOH solution. Then the molarity of the resulting solution is

320mg of sample of magnesium having a coating of its oxide required 20mL of 0.1M hydrochloric acid for the complete neutralisation of the latter. The composition of the sample is :

250ml of a 1.5M solution of sulphuric acid is diluted by adding 5L of water. what is the molarity of the diluted solution?