Chemistry · 2021
JEE Main · 4 August 2021, Shift 2 · Q51
The amount of potassium chlorate ( KClO_3) that needs to be completely decomposed in order to liberate 40 dm^3 of oxygen gas at STP is \ \ \ \ g.…
The amount of potassium chlorate $\displaystyle \left(\mathrm{KClO}_3\right)$ that needs to be completely decomposed in order to liberate $\displaystyle 40 \mathrm{dm}^3$ of oxygen gas at STP is \_\_\_\_ g. (Nearest integer)
[Atomic Masses : $\displaystyle \mathrm{K}=39.0 \mathrm{u} ; \mathrm{Cl}=35.5 \mathrm{u} ; \mathrm{O}=16.0 \mathrm{u}, \mathrm{R}=0.0831 \mathrm{~L}$ bar $\displaystyle \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
Assume oxygen is an ideal gas at STP; at STP molar volume of an ideal gas is $\displaystyle 22.7 \mathrm{~L} \mathrm{~mol}^{-1}$ ]
Official answer
144
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.