Question

When $4.215g$ of metallic carbonate was heated in a hard glass tube, the $C{O_2}$ evolved was found to be measured $1336mL$ at $27^\circ C$ and $700mm$ pressure. If $Xg$ is the equivalent weight of the metal, then find the value of $1000X$.

Answer 1

Hint: To solve this question, to find equivalent weight of metal first we need to find the given weight for carbon dioxide using an ideal gas equation and then we will equate the number of moles of metal carbonate and carbon dioxide.

Complete step by step answer:
As we know that the equation of formation of carbon dioxide can be written as:
$MC{O_3} \to MO + C{O_2}$
Where $M$ is the metal.
To find an equivalent weight of metal ,we can equate moles of metal carbonate and carbon dioxide because we know that one mole of metal carbonate will produce one mole of carbon dioxide. To find moles of carbon dioxide, we need to find its given weight by using the ideal gas equation.
As we know, ideal gas equation can be written as:
$PV = nRT$
Where, $P = $ pressure , $V = $ volume and $T = $temperature, $n = $number of moles, $R = $ real gas constant.
Also, $R = 0.0821L - atm/mol - K$ and $n = \dfrac{w}{m}$; where $w = $given mass, $m = $molecular mass.
So, the ideal gas equation becomes:
$PV = \dfrac{w}{m}RT$ $ - (1)$
Now,
Given temperature for carbon dioxide $ = T = 27 + 273 = 300K$ (here we did conversion of degree Celsius temperature to kelvin temperature.
Pressure of carbon dioxide $ = P = 700mm$ (given)
Volume of carbon dioxide $ = V = 1336mL$ (given)
To convert pressure in the atmosphere we will divide it by $760$ and to convert volume into litre we will divide volume by $1000$. Also, molecular mass($m$) of carbon dioxide is $44$($12g$ of one carbon and $16g$ of two oxygens). Thus, by putting all values in equation $ - (1)$ we get:
$\dfrac{{700}}{{760}} \times \dfrac{{1336}}{{1000}} = \dfrac{w}{{44}} \times 0.0821 \times 300$
$w = 2.198g$.
Thus, the weight of carbon dioxide is $2.198g$. Now, as we know number of moles of metal carbonate and carbon dioxide are equal so by equating these we get:
${n_{MC{O_3}}} = {n_{C{O_2}}}$ $ - (2)$
Here, molecular mass of any metal carbonate can be given as : $2E + 60$ where $E = $equivalent weight of metal and $60$ is the mass of carbonate. Given mass of metal carbonate is $4.215g$. Molecular mass of carbon dioxide is $44$ and its given mass is $2.198g$. Now, by putting all values in equation $ - (2)$ we get:
$\dfrac{{4.215}}{{2E + 60}} = \dfrac{{2.198}}{{44}}$
$E = 12.188g$
Hence, the equivalent weight of the metal is $12.188g$ which is equal to $X$.
Hence, $1000X = 12188g$

Note:
In this question, we need to remember that molecular mass of any carbonate is given as $2E + 60$. It is the same either there is one metal in metal carbonate ($MC{O_3}$) or there are two metals in metal carbonate(${M_2}C{O_3}$).