Question

A sample of chalk contained as an impurity, a form of clay which loses \[14.5\% \] its mass of water on prolonged heating. \[5g\] of chalk on heating shows a loss of \[1.507g\] in mass (due to an evolution of \[C{O_2}\] and water). The \[\% \] of chalk in the sample is: (to the closest integer)

Answer 1

Hint:The question says that chalk contains impurities. We know that chalk contains impurities in the form of clay and calcium carbonate \[CaC{O_3}\]. On heating, there is a loss of mass due to the evolution of gases and water. To calculate the \[\% \] chalk in the sample we will calculate the \[\% CaC{O_3}\].

Formula Used:\[\% CaC{O_3} = \dfrac{{{W_{CaC{O_3}}}}}{W} \times 100\]
Where \[{W_{CaC{O_3}}}\] is the mass of calcium carbonate in the sample and \[W\] is the total mass of the sample containing calcium carbonate and clay.

Complete step-by-step solution:First, we will understand the given conditions and process given in the question step by step. As we have given the sample of chalk which contains clay as an impurity and calcium carbonate. So, the question says that \[5g\] chalk on heating shows a loss of \[1.507g\] mass. So we can consider that in the sample there is \[Xg\] clay and \[Yg\] calcium carbonate. So we can conclude that,
 \[X + Y = 5\] \[ - \left( 1 \right)\]
 Now when \[5g\] of chalk is heated the following reactions take place,
\[CaC{O_3}\xrightarrow{\Delta }CaO + C{O_2};clay\xrightarrow{\Delta }loses + water\]
Now according to the question, we can calculate the mass loss by \[Xg\] clay in form of \[{H_2}O = \dfrac{{14.5}}{{100}} \times X\]. Similarly, we can also calculate the mass loss by \[Yg\] calcium carbonate in form of \[C{O_2} = \dfrac{{44}}{{100}} \times Y\]. We have given the total loss as \[1.507g\]. So we can conclude that, \[\dfrac{{14.5 \times X}}{{100}} + \dfrac{{44 \times Y}}{{100}} = 1.507\] \[ - \left( 2 \right)\]
By solving equations \[ - \left( 1 \right)\]and \[ - \left( 2 \right)\] we get, \[X = 2.349g,Y = 2.651g\]. Now we have calculated\[{W_{CaC{O_3}}} = Y = 2.651g,W = 5g\]. So using the formula, \[\% CaC{O_3} = \dfrac{{{W_{CaC{O_3}}}}}{W} \times 100\] we can calculate the \[\% \] chalk in the sample as \[\% CaC{O_3} = \dfrac{{2.651}}{5} \times 100 = 53.02 \approx 53\].
So we can conclude that \[53\% \] chalk is present in the sample.

Note:Chalk is a non-clastic carbonate sedimentary rock form of limestone composed of the mineral calcite. Chalk is composed of the shells of such minute organisms such as coccoliths, foraminifera, and rhodoliths. The purest chalk contains \[99\% CaC{O_3}\] the form of mineral calcite.