The chloride of a metal contains 71% chlorine by weight and the vapour density of it is 50. The atomic weight of the metal will be (a)29 (b)58 (c)35.5 (d)71 correct answer is option A can you explain the answer?

Question

Answer

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

Answer

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.

Answer

=>   Vapor density = Molar mass of the gas/molar mass of hydrogen.  =>  molar mass of gas = molar mass of Hydrogen(= 2 ) × Vapor density of gas =>The molar mass of this chloride metal = 2 x 50                                                              =  100   =>In the metal chloride 71% is chlorine Then, that means 29% mass is the metal =>In 100g of this metal the mass of the metal = 29/100 × 100g = 29g =>moles = mass / molar mass            = 71/35.5            = 2    =>the empirical formula of the metal chloride be: =>XCl₂   =>empirical formula weight = x (molar mass × moles = mass)+(35.2 × 2)  =>29  + 71 =>29+ 71/100 = 100/100 =>formula ratio = 1 Therefore the formula of the metal chloride is XCl₂ =>molar mass x moles = mass                                      =>molar mass = mass/moles                      = 29/1 =>therefore molar mass = 29 =>The atomic weight of the metal = 29

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