116 mg of a compound on vaporisation in a Vi | vedclass

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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

$58$

Vedclass · Answer

(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 $= \frac{	ext{Mass}}{	ext{Moles}} = \frac{116 	imes 10^{-3} \ g}{0.002 \ mol} = 58 \ g/mol$.

Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

$116 \ mg$ 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

Similar Questions

Upon fully dissolving $2.0 \ g$ of a metal in sulphuric acid,$6.8 \ g$ of the metal sulphate is formed. The equivalent weight of the metal is $...... \ g$.

Calculate the atomic mass (average) of chlorine using the following data:

Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$

(in $u$)

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