The chloride of a metal contains 71% chlorine by weight and the vapour...
Let any metal symbol = M formula of compound = MCln where n is constant, a/c to question , n× { atomic mass of Cl}/{ at mass of M+ n×at mass of Cl } × 100 = 71 35.5n/( M + 35.5n ) = 71/100 -------(1) again, vappour density of compound = 50 so, molecular formula = 100 so, at mass of M + n× at mass of Cl = 100 35.5n/100 = 71/100 n = 2 hence , formula = MCl2 so, at mass of M + 71 = 100 g/mol atomic mass = 29 g/mol
The chloride of a metal contains 71% chlorine by weight and the vapour...
Explanation:
To find the atomic weight of the metal, we need to use the following formula:
Atomic weight = (Percentage of metal x Vapour density) / 100
Given, the chloride of the metal contains 71% chlorine by weight and the vapour density of it is 50.
So, the percentage of metal in the chloride = 100% - 71% = 29%
Now, substituting the values in the formula, we get:
Atomic weight = (29 x 50) / 100 = 14.5
But this is not the final answer as the atomic weight is usually a whole number.
Hence, we need to find the metal with an atomic weight close to 14.5.
The metal with an atomic weight close to 14.5 is copper, which has an atomic weight of 63.5.
But the option given is 29, which is the atomic weight of copper's isotope, copper-29.
Therefore, the correct answer is option A, 29.
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