1.

Calculate the volume occupied by 6.02 × 1025 molecules of oxygen at STP.

Answer»

Volume occupied by 1 mole of oxygen gas at STP = 22.4 l

i.e., Volume occupied by 6.02×1023 molecules of oxygen gas at STP = 22.4 l

Hence the volume occupied by 6.02 × 1025 molecules of oxygen gas at STP = \(\frac{22.4\times6.02\times10^{25}}{6.02\times10^{23}}\) =  2240 l.



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