Questions & Answers

Question

Answers

A ) 26.95

B ) 30

C ) 29.50

D ) 29.90

Answer
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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 \[26.95{\text{ }}g/mol\] .

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