Answer
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Hint: As per the definition of molar mass it is the ratio mass of a substance to the moles of the substance. Using the given weight and moles of each compound calculate the molar mass of each compound. Assume the atomic weight of X and Y as x and y respectively, set up the equation for molar mass of \[{\text{X}}{{\text{Y}}_{\text{2}}}\] and \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] in terms of x and y. Using the calculated molar masses and these two equations calculate the atomic weights of X and Y.
Complete answer:
Let us assume that the atomic weight of the element X is ‘x’ and the atomic weight of the element Y is ‘y’.
As we know the molar mass of the compound is the sum of the atomic weights.
So, the molar mass of \[{\text{X}}{{\text{Y}}_{\text{2}}}\] = 1 (atomic weight of X) + 2 (atomic weight of Y)
\[{\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = {\text{ x + 2y}}\] - (Eq 1)
We can also set up the equation for molar mass of \[{\text{X}}{{\text{Y}}_{\text{2}}}\] using the given moles and weight of \[{\text{X}}{{\text{Y}}_{\text{2}}}\].
We have given 0.1 mol of \[{\text{X}}{{\text{Y}}_{\text{2}}}\]weighs 10 g
\[{\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = \dfrac{{{\text{weight}}}}{{{\text{mole}}}}\]
\[\Rightarrow {\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = \dfrac{{10{\text{ g}}}}{{0.1{\text{ mol}}}}\]
\[\Rightarrow {\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = {\text{ 100 g/mol }}\] - (Eq 2)
So, now by relating equation 1 and 2 we can say that,
\[{\text{ x + 2y = 100}}\] - (Eq 3)
We can rearrange this equation for x as follows:
\[\Rightarrow {\text{ x = 100 - 2y}}\] - (Eq 4)
Similarly, now we can set up the equation for molar mass of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] as follows:
Molar mass of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] = 3 (atomic weight of X) + 2 (atomic weight of Y)
\[ {\text{Molar mass }}{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}} = {\text{ 3x + 2y}}\] - (Eq 5)
We can also set up the equation for molar mass of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] using the given moles and weight of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\].
0.05 mol of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] weighs 9 g
\[\Rightarrow {\text{Molar mass }}{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}} = \dfrac{{{\text{9 g}}}}{{0.05{\text{ mol}}}}\]
\[\Rightarrow {\text{Molar mass }}{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}} = 18{\text{0 g/mol}}\] - (Eq 6)
So, now by relating equation 5 and 6 we can say
\[{\text{3x + 2y = 180}}\] - (Eq 7)
Now, substituting the value of x from equation (4 )to equation (7) we can solve the equation for y.
\[\Rightarrow {\text{3(100 - 2y) + 2y = 180}}\]
\[\Rightarrow {\text{300 - 6y + 2y = 180}}\]
\[\Rightarrow {\text{300 - 4y = 180}}\]
\[\Rightarrow y = 30\]
Thus, the value of y is 30.
Now, to calculate the value of x we can substitute this value of y in equation (4).
\[\Rightarrow {\text{ x = 100 - 2(30) = 40}}\]
Thus, the value of x is 40.
Hence, the atomic weight of X is 40 and the atomic weight of Y is 30.
Thus, the correct option is (C) 40, 30
Note: We can also solve this problem by calculating the value of y first. However, for that, we have to rearrange equation 3 for y. But this rearrangement will give us a complicated equation which will make the whole solution complicated.
Complete answer:
Let us assume that the atomic weight of the element X is ‘x’ and the atomic weight of the element Y is ‘y’.
As we know the molar mass of the compound is the sum of the atomic weights.
So, the molar mass of \[{\text{X}}{{\text{Y}}_{\text{2}}}\] = 1 (atomic weight of X) + 2 (atomic weight of Y)
\[{\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = {\text{ x + 2y}}\] - (Eq 1)
We can also set up the equation for molar mass of \[{\text{X}}{{\text{Y}}_{\text{2}}}\] using the given moles and weight of \[{\text{X}}{{\text{Y}}_{\text{2}}}\].
We have given 0.1 mol of \[{\text{X}}{{\text{Y}}_{\text{2}}}\]weighs 10 g
\[{\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = \dfrac{{{\text{weight}}}}{{{\text{mole}}}}\]
\[\Rightarrow {\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = \dfrac{{10{\text{ g}}}}{{0.1{\text{ mol}}}}\]
\[\Rightarrow {\text{Molar mass X}}{{\text{Y}}_{\text{2}}} = {\text{ 100 g/mol }}\] - (Eq 2)
So, now by relating equation 1 and 2 we can say that,
\[{\text{ x + 2y = 100}}\] - (Eq 3)
We can rearrange this equation for x as follows:
\[\Rightarrow {\text{ x = 100 - 2y}}\] - (Eq 4)
Similarly, now we can set up the equation for molar mass of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] as follows:
Molar mass of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] = 3 (atomic weight of X) + 2 (atomic weight of Y)
\[ {\text{Molar mass }}{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}} = {\text{ 3x + 2y}}\] - (Eq 5)
We can also set up the equation for molar mass of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] using the given moles and weight of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\].
0.05 mol of \[{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}}\] weighs 9 g
\[\Rightarrow {\text{Molar mass }}{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}} = \dfrac{{{\text{9 g}}}}{{0.05{\text{ mol}}}}\]
\[\Rightarrow {\text{Molar mass }}{{\text{X}}_{\text{3}}}{{\text{Y}}_{\text{2}}} = 18{\text{0 g/mol}}\] - (Eq 6)
So, now by relating equation 5 and 6 we can say
\[{\text{3x + 2y = 180}}\] - (Eq 7)
Now, substituting the value of x from equation (4 )to equation (7) we can solve the equation for y.
\[\Rightarrow {\text{3(100 - 2y) + 2y = 180}}\]
\[\Rightarrow {\text{300 - 6y + 2y = 180}}\]
\[\Rightarrow {\text{300 - 4y = 180}}\]
\[\Rightarrow y = 30\]
Thus, the value of y is 30.
Now, to calculate the value of x we can substitute this value of y in equation (4).
\[\Rightarrow {\text{ x = 100 - 2(30) = 40}}\]
Thus, the value of x is 40.
Hence, the atomic weight of X is 40 and the atomic weight of Y is 30.
Thus, the correct option is (C) 40, 30
Note: We can also solve this problem by calculating the value of y first. However, for that, we have to rearrange equation 3 for y. But this rearrangement will give us a complicated equation which will make the whole solution complicated.
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