Question

A sample of \[{{C}_{3}}{{H}_{4}}\] and \[{{C}_{5}}{{H}_{12}}\] gas has \[d%\], by mass of \[{{C}_{3}}{{H}_{4}}\]. If the average molar mass of the sample is \[200/3{ }gm\], then the value of \[d\] is:
A.\[0.1\]
B.\[10\]
C.\[1/6\]
D.\[20\]

Answer 1

Hint:The average molar mass of a compound is the sum of the molar masses of each compound present in a solution, along with their mass percentage.
-Mole fraction of a component represents the number of molecules of a specific component in a mixture divided by the total number of moles in the given mixture.

Complete step by step answer:
In the field of chemistry, the molar mass of a chemical compound can be defined as the mass of a sample of that compound divided by the amount of substance taken in that sample, which is measured in moles. The property of molar mass is a bulk property and not a molecular property of a substance. The property of molar mass can be calculated as an average of many instances of the compound or a molecule, which often differs in mass because of the presence of isotopes. The molar mass is appropriate for the conversion between the amounts of a substance for bulk quantities and mass of a substance.
 A very commonly used synonym of molar mass is the molecular weight, which is particularly used for molecular compounds.
In the given question, we know that Molar mass of \[{{C}_{3}}{{H}_{4}}\]\[:40g\]
And Molar mass \[{{C}_{5}}{{H}_{12}}\]is\[~72\]
It is given that the total sample contains both of these sample and the average molar mass of the sample is \[200/3{ }gm\], so based on this information, we will add the molar masses of the compounds, and equate it with the average molar mass of the sample. The general formula can be represented in a simple way as,
\[[Molar{ }mass\left( of{ }{{C}_{3}}{{H}_{4}} \right)\times %by{ }mass{ }ratio]{ }+[molar{ }mass{ }of{ }\left( {{C}_{5}}{{H}_{12}} \right)\times \left( 1-%by{ }mass{ }ratio \right)]=avg.molarmassOf{ }sample\] now putting the known values we get,
\[40d+72\left( 1-d \right)=\dfrac{200}{3}\]
Now we will solve the following equation, mass percentage of the \[{{C}_{3}}{{H}_{4}}\],
\[40d+72-72d=\dfrac{200}{3}\]
Now we will keep the unknown value at the right hand side of the equation and the known value to the left hand side of the equation in order to solve it more easily,
\[72-\dfrac{200}{3}=32d\]
\[\dfrac{16}{3}{ =}32d\]
Hence the value of d is,
\[d=\dfrac{1}{6}\]​
Since the calculated value came out to be \[1/6\], so we the correct answer would be option C.

Note:
-Percentage by mass of a component present in the mixture tells us about the amount of that component which is present in the mixture.
-If the mixture contains only two components, as in the case of the given question, we can always calculate the value of mass percentage by using the average molar mass of the sample of mixture.