100mL each of `1N H_(2)O_(2)` and `11.2VH_(2)O_(2)` solution are mixed, then the final solution is equivalent to:
(Assume 1 mole od an ideal gas occupies 22.4L at STP)

The volume strength of 0.75 M H_(2)O_(2) solution is : (Assume 1 mole of an ideal gas occupies 22.4L at STP)

20 ml of 20 vol H_(2)O_(2) solution is diluted to 80 ml. The final volume strength of solution is

56V,500ml H_2O_2 solution is kept in an open contaier due to which some H_2O_2 is decomposed and evolves 8gm O_2 . Simultaneously some H_2O also vapouries. Due to all these changes ,final volume is reduced by 20% . Find final volume strength of H_2O_2(aq) . Assume 1 mole of an ideal gas occupies 22.4 L at STP)

When 100 ml of 10M solution of H_(2)SO_(4) and 100ml of 1M solution of NaOH are mixed the resulting solution will be

If 100 mL of 1 N H_(2)SO_(4) is mixed with 100 mL of 1 M NaOH solution. The resulting solution will be

If 100 mL of 1 N H_(2)SO_(4) is mixed with 100 mL of 1 M NaOH solution. The resulting solution will be

Bottle (A) contains 320 mL of H_(2)O_(2) solution and labelled with 10 V H_(2)O_(2) and bottle (B) contains 80 mL H_(2)O_(2) having molarity 2.5 M. Content of bottle (A) and bottle (B) are mixed and solution is filled in bottle (C). Select the correct label for bottle (C) in terms of volume strength and g/litre. (Assume 1 mole of an ideal gas occupies 22.4 L at STP)