0.583 g of acetone, when vapourised, formed 225 $c m^{3}$ of vapours at S.T.P. Calculate the gram molecular mass of the gas. [22.4 L of any gas at S.T.P. = 1 g.mol. mass of gas]

56 g

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58.04 g

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86.44 g

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0.0508 g

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Solution

The correct option is B
58.04 g

225 $c m^{3}$ of acetone at S.T.P. weighs = 0.583 g

22400 $c m^{3}$ of acetone at stp will weigh = $\frac{0.583 \times 22400}{225}$ = 58.04 g.

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