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(C) Given mass of the compound $= 116 \ mg = 116 	imes 10^{-3} \ g$.Volume of air displaced at $S.T.P. = 44.8 \ mL = 44.8 	imes 10^{-3} \ L$.At $S.T

Vedclass · Accepted Answer

$58$

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(C) Given mass of the compound $= 116 \ mg = 116 	imes 10^{-3} \ g$.Volume of air displaced at $S.T.P. = 44.8 \ mL = 44.8 	imes 10^{-3} \ L$.At $S.T.P.$,$1 \ mole$ of any gas occupies $22.4 \ L$ or $22400 \ mL$.Number of moles of the compound $= \frac{	ext{Volume at S.T.P. (mL)}}{22400 \ mL/mol} = \frac{44.8}{22400} = 0.002 \ mol$.Molecular weight $= \frac{	ext{Mass}}{	ext{Moles}} = \frac{116 	imes 10^{-3} \ g}{0.002 \ mol} = 58 \ g/mol$.

$116 \ mg$ of a compound on vaporisation in a Victor Meyer's apparatus displaces $44.8 \ mL$ of air measured at $S.T.P.$ The molecular weight of the compound is

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