Decreasing order (first having highest and then other following it) of mass of pure NaOH in each of the aqueous solution
(P) 50 gm of `40%(w//w)` NaOH
(Q) 50 gm of `50%(w//V)` NaOH `[d_("soln.")=1.2gm//ml]`
(R) 50 gm of 20 M NaOH `[d_("soln").=1gm//ml]`

Calclate pH of a resultant solution of 25 mL of 0.1 M HCl, 50 mL of 0.02 M HNO_(3) and 25 mL of 0.1M NaOH

100mL of 10% NaOH(w/V) is added to 100 mL of 10% HCI(w/V) . The nature of resultant solution is :

The density of 2.0 M solution of solute is 1.2 gm/ml. If the molecular mass of solute is 100 gm "mol"^(-1) then the molality of the solution is

A man starts from the point P(−3,4) and reaches at point Q(0,1) touches x-axis at point R such that PR+RQ is least then position of point R is (a) (3/5,0) (b) (-3/5,0) (c) (-2/5,0) (d) (-2,0)

A mixture of H_(2)SO_(4) and H_(2)C_(2)O_(4) (oxalic acid ) and some inert impurity weighing 3.185 g was dessolved in water and the solution made up to 1litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO_(4) solution for complete reaction. The mass % of H_(2)SO_(4) in the mixture was:

How many millilitres of 0.1 M H_2SO_4 must be added to 50mL of 0.1 M NaOH to give a solution that has a concentration of 0.05 M in H_2SO_4 ?

To a 10mL, 1M aqueous solution of Br_(2) , excess of NaOH is added so that all Br_(2) is disproportionated to Br^(-) and BrO_(3)^(-) . The resulting solution is free from Br^(-) , by extraction and excess of OH^(-) neutralised by acidifying the solution. The resulting solution is suffcient to react with 2 g of impure CaC_(2)O_(4) (M= 128g/mol) sample. The % purity of oxalate sample is :

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is :

A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 mL. Of 0.1 M NaOH has been added. Now 20 mL of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0 What will be the pH of the solution obtained by mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA?

Calculate [H^(+)] at equivalent point between titration of 0.1 M, 25 mL of weak acid HA (K_(a) (HA)=10^(-5)) with 0.05 M, 50ml of NaOH solution: