
A mixture of gas X (mol. wt. 16) and gas Y (mol. wt. 28) in the mole ratio \[{\text{a:b}}\] has a mean molecular weight 20. What would be the mean molecular weight if the gases are mixed in the ratio \[{\text{b:a}}\] under identical conditions (gases are non reacting)?
A.24
B.20
C.26
D.40
Answer
565.5k+ views
Hint: We are given that the gases are non reacting. This means that the reaction does not occur between gas X and gas Y. Thus, the number of moles will not be affected as the reaction does not occur.
Complete step by step answer:
We are given that a mixture of gas X (mol. wt. 16) and gas Y (mol. wt. 28) in the mole ratio \[{\text{a:b}}\] has a mean molecular weight 20.
Thus, the equation of mean molecular weight can be written as follows:
$\dfrac{{16{\text{a}} + 28{\text{b}}}}{{{\text{a}} + {\text{b}}}} = 20$
$16{\text{a}} + 28{\text{b}} = 20{\text{a}} + 20{\text{b}}$
$8{\text{b}} = 4{\text{a}}$
$2{\text{b}} = {\text{a}}$
Now, calculate the mean molecular mass when the mixture of gas X (mol. wt. 16) and gas Y (mol. wt. 28) is in the mole ratio \[{\text{b:a}}\].
Thus,
${\text{Mean molecular weight}} = \dfrac{{16{\text{b}} + 28{\text{a}}}}{{{\text{a}} + {\text{b}}}}$
Substitute ${\text{a}} = {\text{2b}}$ in the above equation. Thus,
${\text{Mean molecular weight}} = \dfrac{{16{\text{b}} + 28\left( {{\text{2b}}} \right)}}{{{\text{2b}} + {\text{b}}}}$
${\text{Mean molecular weight}} = \dfrac{{16{\text{b}} + {\text{56b}}}}{{3{\text{b}}}}$
${\text{Mean molecular weight}} = \dfrac{{\left( {16 + {\text{56}}} \right){\text{b}}}}{{3{\text{b}}}}$
${\text{Mean molecular weight}} = \dfrac{{72}}{3}$
${\text{Mean molecular weight}} = 24$
Thus, the mean molecular weight if the gases are mixed in the ratio \[{\text{b:a}}\] under identical conditions is 24.
Thus, the correct option is option (A).
Note:
The sum of the atomic weights of all the atoms that make up the molecular formula of a compound is known as the molecular weight of the compound. The molecular weight per unit mole is known as the mean molecular weight. The general formula to calculate mean molecular mass is as follows:
${\text{Mean molecular weight}} = \dfrac{{\sum {N_i}{M_i}}}{{\sum {N_i}}}$
Where $M$ is the molecular weight and,
$N$ is the number.
Complete step by step answer:
We are given that a mixture of gas X (mol. wt. 16) and gas Y (mol. wt. 28) in the mole ratio \[{\text{a:b}}\] has a mean molecular weight 20.
Thus, the equation of mean molecular weight can be written as follows:
$\dfrac{{16{\text{a}} + 28{\text{b}}}}{{{\text{a}} + {\text{b}}}} = 20$
$16{\text{a}} + 28{\text{b}} = 20{\text{a}} + 20{\text{b}}$
$8{\text{b}} = 4{\text{a}}$
$2{\text{b}} = {\text{a}}$
Now, calculate the mean molecular mass when the mixture of gas X (mol. wt. 16) and gas Y (mol. wt. 28) is in the mole ratio \[{\text{b:a}}\].
Thus,
${\text{Mean molecular weight}} = \dfrac{{16{\text{b}} + 28{\text{a}}}}{{{\text{a}} + {\text{b}}}}$
Substitute ${\text{a}} = {\text{2b}}$ in the above equation. Thus,
${\text{Mean molecular weight}} = \dfrac{{16{\text{b}} + 28\left( {{\text{2b}}} \right)}}{{{\text{2b}} + {\text{b}}}}$
${\text{Mean molecular weight}} = \dfrac{{16{\text{b}} + {\text{56b}}}}{{3{\text{b}}}}$
${\text{Mean molecular weight}} = \dfrac{{\left( {16 + {\text{56}}} \right){\text{b}}}}{{3{\text{b}}}}$
${\text{Mean molecular weight}} = \dfrac{{72}}{3}$
${\text{Mean molecular weight}} = 24$
Thus, the mean molecular weight if the gases are mixed in the ratio \[{\text{b:a}}\] under identical conditions is 24.
Thus, the correct option is option (A).
Note:
The sum of the atomic weights of all the atoms that make up the molecular formula of a compound is known as the molecular weight of the compound. The molecular weight per unit mole is known as the mean molecular weight. The general formula to calculate mean molecular mass is as follows:
${\text{Mean molecular weight}} = \dfrac{{\sum {N_i}{M_i}}}{{\sum {N_i}}}$
Where $M$ is the molecular weight and,
$N$ is the number.
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