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