JEE MAIN - Chemistry (2020 - 3rd September Morning Slot - No. 9)
The volume strength of 8.9 M H2O2
solution calculated at 273 K and 1 atm is ______. (R = 0.0821 L
atm K-1 mol-1) (rounded off ot the nearest integer)
Answer
100
Explanation
Volume strength of H2O2 at 1 atm
273 kelvin
= M × 11.2 = 8.9 × 11.2 = 99.68 $$ \simeq $$ 100
= M × 11.2 = 8.9 × 11.2 = 99.68 $$ \simeq $$ 100
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