The empirical formula of a compound is $A B_{2}$ . If its empirical formula weight is 2/3 times of its vapour density, calculate the molecular formula of the compound.

A_{3} B_{6}

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A_{2} B_{4}

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A_{6} B_{12}

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A_{4} B_{8}

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Solution

The correct option is A $A_{3} B_{6}$
Given that Emperical weight = $\frac{2}{3}$ (V.D)
Now Molecular weight = 2(V.D)
Therefore n = $\frac{2 \times V . D}{\frac{2}{3} V . D}$ = 3
Therefore molecular formula = $(A B_{2})_{3}$ = $A_{3} B_{6}$

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The empirical formula of a compound is AB2. If its empirical formula weight is 2/3 times of its vapour density, calculate the molecular formula of the compound.

The correct option is A A3B6Given that Emperical weight = 23(V.D)Now Molecular weight = 2(V.D)Therefore n = 2×V.D23V.D = 3Therefore molecular formula = (AB2)3 = A3B6

The empirical formula of a compound is $A B_{2}$ . If its empirical formula weight is 2/3 times of its vapour density, calculate the molecular formula of the compound.

The correct option is A $A_{3} B_{6}$
Given that Emperical weight = $\frac{2}{3}$ (V.D)
Now Molecular weight = 2(V.D)
Therefore n = $\frac{2 \times V . D}{\frac{2}{3} V . D}$ = 3
Therefore molecular formula = $(A B_{2})_{3}$ = $A_{3} B_{6}$