Commerical sample of `H_2 O_2` is labeled as 10 V. its % strength is nearly

Answer

3

Explanation

Applying the formula, we have,
Volume strength `=5.6 xx ("Percentage strength")/("Eq. Wt of" H_(2)O_(2)) xx 10`
Percentage strength `= 17/5.6 = 3.03 = 3`