Question

The element X and Y form compounds having molecular formula \[X{Y_2}\] and \[X{Y_4}\]. When dissolved in a 20 gm of benzene, 1 gm \[X{Y_2}\] lowers the freezing point by \[{2.3^ \circ }\,\], whereas 1 gm of \[X{Y_4}\] lowers the freezing point by $1.3{\,^ \circ }\,C$. The molal depression constant for benzene is 5.1. Calculate the atomic masses of X and Y.
A. \[x = 25.6,y = 42.6\]
B. $x = 42.6,y = 25.6$
C. \[x = 35.89,y = 54.9\]
D. $x = 54.9,y = 35.89$

Answer 1

Hint: In this, we have to calculate the molar masses of element X and Y. According to the present definition, the atomic mass of an element is the average relative mass of an atom of the element as compared to the mass of $^{12}C$ an atom taken as 12 units. The freezing point of a substance is defined as the temperature at which it just changes into the solid-state and both solid and liquid co-exist and has the same vapor pressure.

Complete step by step answer:
Now we will calculate the given numerical.
First we will calculate the molecular masses of both the compounds.
For compound \[X{Y_2}\]
Weight of \[X{Y_2}\] $\left( {{W_{X{Y_2}}}} \right)$ is 1 gm
Depression in freezing point of \[X{Y_2}\] $\left( {\Delta {T_f}} \right)$ is \[{2.3^ \circ }\,\]
Weight of benzene (solvent) is 20 gm
Molal elevation constant $\left( {{K_f}} \right)$ for benzene is 5.1
Now we will calculate the molecular mass of \[X{Y_2}\]
${M_{X{Y_2}}} = \dfrac{{{K_f} \times {W_{X{Y_2}}} \times 1000}}{{\Delta {T_f} \times {W_1}}}$
$ \Rightarrow {M_{X{Y_2}}} = \dfrac{{5.1 \times 1 \times 1000}}{{2.3 \times 20}}$
$ \Rightarrow {M_{X{Y_2}}} = 110.86g/mol$
(After putting the values and calculating we get the molecular mass of the compound \[X{Y_2}\] is 110.86 g/mol.)
For compound \[X{Y_4}\] , Weight of \[X{Y_4}\] $\left( {{W_{X{Y_4}}}} \right)$ is 1 gm
Depression in freezing point of \[X{Y_4}\] $\left( {\Delta {T_f}} \right)$ is $1.3{\,^ \circ }\,C$
Now, we will calculate the molecular mass of \[X{Y_4}\]
${M_{X{Y_4}}} = \dfrac{{{K_f} \times {W_{X{Y_4}}} \times 1000}}{{\Delta {T_f} \times {W_1}}}$
$ \Rightarrow {M_{X{Y_4}}} = \dfrac{{5.1 \times 1 \times 1000}}{{1.3 \times 20}}$
$ \Rightarrow {M_{X{Y_4}}} = 196.15g/mol$
(After putting the values and calculating we get the molecular mass of the compound \[X{Y_4}\] is
196.15 g/mol.)
Let the atomic masses of the element X and Y be x and y
${M_{X{Y_2}}} = x + 2y = 110.86\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 1 \right)$
${M_{X{Y_4}}} = x + 4y = 196.15\,\,\,\,\,\,\,\,\,\,\,\,\, \to (2)$
Subtracting equation (1) from equation (2) we get
$ \Rightarrow x + 4y - \left( {x + 2y} \right) = 196.15 - 110.86$
$ \Rightarrow 2y = 85.29$
$ \Rightarrow y = 42.6$
Putting the value of y in equation (1) we get,
$ \Rightarrow x + 2 \times 42.6 = 110.86$
$ \Rightarrow x = 110.86 - 85.2$
$ \Rightarrow x = 25.6$
Therefore, the atomic masses of the element X and Y are 25.6 and 42.6

So, the correct answer is Option A.

Note: The freezing point of a solution is always less than that of its pure solvent. The vapor pressure of the solution will become equal to that of the pure solid solvent only at a low temperature so that it may start freezing. This is the reason why the freezing point of a solution is less than that of its pure solvent.