Question

Two oxides of metals A and B are isomorphous. The metal A whose atomic mass is 52, forms a chloride whose vapour density is 79. The oxide of the metal B contains 47.1% oxygen. Calculate the atomic mass of B.
A ) 26.95
B ) 30
C ) 29.50
D ) 29.90

Answer 1

Hint: Vapour density is the density of a gas relative to the density of hydrogen atom under identical conditions of temperature and pressure. From the vapour density, calculate the molecular weight of chloride of metal A. From the molecular weight of chloride of metal A and the atomic mass of metal A and chlorine, determine the valence of metal A. Then determine the molecular formula of metals A and B. Then from the formula for percentage of oxygen in the oxide of metal B, calculate the atomic mass of metal B.

Complete step by step answer:
The atomic mass of metal A is 52. It forms a chloride. Let the molecular formula of the chloride of metal A be \[{\text{MC}}{{\text{l}}_x}\]. Here M represents the metal.
The atomic mass of chlorine is 35.5. Calculate the molecular weight of chloride of metal A \[52 + \left( {35.5} \right)X\]
The vapour density of the chloride of metal A is 79. Molecular weight is twice the vapour density. Calculate the molecular weight of chloride of metal A.
\[2 \times 79 = 158\]
\[
  52 + \left( {35.5} \right)X = 158 \\
  35.5X = 158 - 52 = 106 \\
  X = \dfrac{{106}}{{35.5}} \\
  X = 3 \\
 \]
Hence, the valence of metal is +3. It will form an oxide \[{{\text{M}}_2}{{\text{O}}_3}\]. The valence of oxygen is -2.
Two oxides of metals A and B are isomorphous. Hence, the oxide of metal B will have a molecular formula \[{{\text{M}}_2}{{\text{O}}_3}\] . It contains 47.1% oxygen.
Let B g/mol be the atomic mass of metal B.
Use the formula for the percent of metal oxide and calculate atomic mass of metal B.
\[
\% {{\text{O}}_2}{\text{ = 100 }} \times {\text{ }}\dfrac{{3\left( {16} \right)}}{{2{\text{B}} + 3\left( {16} \right)}}{\text{ }} \\
{\text{47}}{\text{.1 = 100 }} \times {\text{ }}\dfrac{{48}}{{2{\text{B}} + 48}} \\
2{\text{B}} + 48 = {\text{100 }} \times {\text{ }}\dfrac{{48}}{{47.1}} \\
2{\text{B}} + 48 = 101.9 \\
2{\text{B}} = 101.9 - 48 \\
2{\text{B}} = 53.9 \\
{\text{B}} = \dfrac{{53.9}}{2} \\
{\text{B}} = 26.95 \\
 \]
Hence, the atomic mass of metal B is \[26.95{\text{ }}g/mol\] .

Thus option A ) is the correct answer.

Note: Molecular weight is twice the vapour density. Isomorphous substances have similar chemical formulas and similar crystal structure. In the chemical formula of metal oxide, the number of oxygen atoms is equal to the valence of metal and the number of metal atoms is equal to the valence of the oxygen atom.