Question

When $4.215\,g$ of a metallic carbonate was heated in a hard glass tube, the $C{O_2}$ evolved was found to measure $1336\,mL$ at ${27^ \circ }C$ and $700\,mm$pressure. The equivalent weight of the metal $($only integer part$)$is
(i) $14$
(ii) $12$
(iii) $11$
(iv) $13$

Answer 1

Hint: Consider the metal carbonate to be ${M_2}C{O_3}$. Upon heating the metal carbonate will evolve$C{O_2}$ gas according to the equation, \[{M_2}C{O_3}\xrightarrow{\Delta }{M_2}O + C{O_2}\]. From the given data calculate the weight of $C{O_2}$ produced. Also according to the equation, the number of moles of ${M_2}C{O_3}$ taken is equal to the number of moles of $C{O_2}$ produced hence calculate the equivalent weight of the metal from here.

Complete step-by-step solution:Let us consider the formula of metallic carbonate is${M_2}C{O_3}$.
When the metallic carbonate is heated it produces the oxide of the metal and carbon dioxide.
${M_2}C{O_3}\xrightarrow{\Delta }{M_2}O + C{O_2}........\left( 1 \right)$
Let us first calculate the weight of $C{O_2}$ obtained from the data given in the question.
Pressure $ = \,700\,mm\, = \,\,\dfrac{{700}}{{760}}\,atm\,\, = \,0.921\,atm$
Temperature $ = \,{27^ \circ }C\, = \,\left( {27 + 273} \right)\,K\, = \,300\,K$
Volume $ = \,1336\,mL\, = \,\,\dfrac{{1336}}{{1000}}\,L\, = \,1.336\,L$
We know, $PV\, = \,nRT........\left( 2 \right)$ where $n\, = \,\dfrac{m}{M}$ $(m$is the weight of $C{O_2}$ produced and $M$is the molecular weight of $C{O_2}\,)$.
Here the value of $R$ is $0.082\,L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}}$ and molecular weight of $C{O_2}\,(M)\, = \,44\,g\,mo{l^{ - 1}}$
Putting the values in equation $\left( 2 \right)$ we get
$0.921\,atm \times 1.336\,L\, = \,\dfrac{m}{{44\,g\,mo{l^{ - 1}}}} \times 0.082\,L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}} \times 300\,K$
$ \Rightarrow \,m\, = \,\dfrac{{0.921\,atm \times 1.336\,L \times 44\,g\,mo{l^{ - 1}}}}{{0.082\,L\,atm\,{K^{ - 1}}\,mo{l^{ - 1}} \times 300\,K}}$
$ \Rightarrow \,m\, = \,2.2\,g$
From equation $\left( 1 \right)$ we can say that the number of moles of ${M_2}C{O_3}$ taken will be equal to the number of moles of $C{O_2}$ produced.
$\therefore \,{\left( n \right)_{{M_2}C{O_3}}} = {\left( n \right)_{C{O_2}}}$
Number of moles of a compound $ = \,\,\dfrac{{Given\,Weight}}{{Molecular\,Weight}}$
$\therefore \,\,\dfrac{{4.215}}{{{M_{{M_2}C{O_3}}}}} = \dfrac{{2.2}}{{44}}.........\left( 3 \right)$, where${M_{{M_2}C{O_3}}}$is the molecular weight of the metallic carbonate.
Let the equivalent weight of the metal in ${M_2}C{O_3}$ be $E$.
Now, ${M_{{M_2}C{O_3}}}\, = \,\left( {2 \times E + 60} \right)$ $(\,\because $Molecular weight of $C{O_3}\, = \,60\,g\,mo{l^{ - 1\,}})$
Putting the value in equation $\left( 3 \right)$ we get
$\dfrac{{4.215}}{{\left( {2E + 60} \right)}} = \dfrac{{2.2}}{{44}}$
$ \Rightarrow \,2E + 60 = \dfrac{{4.215 \times 44}}{{2.2}} = \,84.3$
$ \Rightarrow \,2E\, = \,\left( {84.3 - 60} \right)\, = \,24.3$
$ \Rightarrow \,E\, = \,\dfrac{{24.3}}{2} = \,12.15\, \sim 12$
Therefore, the equivalent weight of the metal is $12$.

Hence the correct answer is (ii) $12$.

Note:The reaction must be properly balanced in order to avoid any error in the calculation of the equivalent weight of the metal. Also the units of pressure, temperature and volume must be converted properly and the $R$ value must be chosen accordingly. Do not mix up the units.