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$2.76\;g$ of silver carbonate on being strongly heated yields a residue weighing.

$(a)\;2.16\;g \\ (b)\;2.48\;g \\(c)\; 2.32\;g \\(d)\;2.64\;g$
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$Ag_2CO_3 \underrightarrow{\Delta} Ag_2O$
In that $Ag_2CO_3$ is $ 2.76\;g ; \qquad Ag_2O$ is $xg$
Molecular mass of $Ag_2CO_3$
$2(108) +(12)+3(16)=276$
Number of moles of $Ag_2CO_3=\large\frac{2.76}{276}=$$0.01 \;mol$
Now 1 mole of $Ag_2CO_3$ releases 1 mole of $w_2$ and 1 mole of $Ag_2O$ remains as residue.
$\therefore $ residue $=0.01 \times $molecular mass of $Ag_2O$
$\qquad= 0.01 \times 232=2.32\;g$
Hence c is the correct answer.
answered Mar 21, 2014 by meena.p

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