$\frac{100}{10}$ N and 0.1 N solution is called

100 $c{m}^3$ of 0.1 N HCl is mixed with 100 $c{m}^3$ of 0.2 N NaOH solution. The resulting solution is

100 mL of $0.5$ N NaOH solution is added to 10 mL of 3 N $H_{2}S{O}_4$ solution and 20 mL of 1 N HCl solution . The mixture is

$100mL$ of $0.2NNaOH$ is mixed with $100$ mL $0.1NHCl$ and the solution is made 1L. The pH of the solution is :

100 mL of 0.2 N HCl solution is added to 100 mL of 0.2 N $AgN{O}_3$ solution. The molarity of nitrate ions in the resulting mixture will be :

100 ml of 0.2 NHCl solution is added to 100 ml of 0.2 N $AgN{O}_3$ solution. The molarity of nitrate ions in the resulting mixture will be

500 mL of $0.1$ N solution of $AgN{O}_3$ is added to 500 mL of a $0.1$ N KCl solution .the concentration of nitrate in the resulting mixture is

100 ml of 0.3 M $HCl$ is mixed with 200 ml of 0.3 M $H_{2}S{O}_4.$ Calculate the final molarity of the $H^{+}$ ions in the resulting solution. Fill your answer by multiplying with 10.

100 ml of 0.2N HCl solution is added to 100 ml of 0.2 N $AgN{O}_3$ solution. The molarity of nitrate ions in the resulting mixture will be

The amount of $KMn{O}_4$ required to prepare 100mL of a $0.1N$ solution in an acidic medium is :

100 c.c of N/10 NaOH solution is mixed with 100 c.c of N/5 HCI solution and the whole volume is made to 1 litre . The pH of the resulting solution will be