The volume of oxygen liberated at STP from 15 mL of 20 volume `H_(2)O_(2)` is

Options

A. 300 mL
B. 150 mL
C. 200 mL
D. 250 mL

Answer

C

Explanation

20-vol `H_(2)O_(2)` means
1 mL of `H_(2)O_(2)` of 20 volume gives =20 mL of `O_(2)` at stp
Thus 15 mL of `H_(2)O_(2)`n of 20 volume gives =15 x20 mL of `O_(2)` at STP
volume of `O_(2)` at STP =300 mL