100 ml of `Na_(2)S_(2)O_(3)` solution is divided into two equal two parts A and B . A part requires 12.5 ml of 0.2 M `I_(2)` solution (acidic medium) and part B is diluted x times and 50 ml of diluted solution requires 5 ml of 0.8 M `I_(2)` solution in basic medium. What is value of x ?

2.48 g of Na_(2)S_(2)O_(3). xH_(2)O is dissolved per litre solution 20 ml of this solution required 10 ml 0.01 M iodine solution. What is value of x ?

2.48 g of Na_(2)S_(2)O_(3). xH_(2)O is dissolved per litre solution 20 ml of this solution required 10 ml 0.01 M iodine solution. What is value of x ?

A 50mL of solution of I_2 is divided into two unequal parts. I part with hypo solution in acidic medium. 15mL of 0.4M hypo was consumed. II part was added with 100mL of 0.3M NaOH solution. Residual base required to mL of 0.3M H_2SO_4 solution for complete neutralization. What was the initial concentration of I_2?

10mL of H_2O_2 solution (volume strength = x ) requires 10mL of N//0.56 MnO_4^(ɵ) solution in acidic medium. Hence x is

10mL of H_2O_2 solution (volume strength = x ) requires 10mL of N//0.56 MnO_4^(ɵ) solution in acidic medium. Hence x is

10mL of H_2O_2 solution (volume strength = x ) requires 10mL of N//0.56 MnO_4^(ɵ) solution in acidic medium. Hence x is

A 150 mL of solution of I_(2) is divided into two unequal parts. I part reacts with hypo solution solution in acidic medium. 15 mL of 0.4 M hypo was consumed. II part was added with 100 mL of 0.3 M NaOH solution. Residual base required 10 mL 0.3 M H_2SO_4 solution for complete neutralization. What was the initial concentration of I_(2) ?