1 mole of chlorine combines with a certain weight of a metal giving 111g of its chloride. The atomic weight of the metal (assuming its valency to be 2) is-
To determine the atomic weight of the metal, we need to use the concept of mole ratios and molar masses.
Given that 1 mole of chlorine combines with a certain weight of the metal to produce 111g of its chloride, we can set up the following equation:
1 mole of chlorine = molar mass of chlorine
x moles of metal = molar mass of metal (assuming the valency of the metal is 2)
To solve for the atomic weight of the metal, we first need to find the molar mass of chlorine. The molar mass of chlorine (Cl) is approximately 35.5 g/mol.
Therefore, 1 mole of chlorine is equal to 35.5 g.
Now, let's calculate the number of moles of chlorine present in 111g of its chloride:
Number of moles = Mass / Molar mass = 111g / 35.5 g/mol ≈ 3.12 moles
Since the valency of the metal is assumed to be 2, the molar ratio between chlorine and the metal is 1:2.
Using the mole ratio, we can say that:
1 mole of chlorine = 2 moles of metal (assuming the valency of the metal is 2)
So, by multiplying the number of moles of chlorine by 2, we can find the number of moles of metal:
Number of moles of metal = 3.12 moles x 2 = 6.24 moles
Finally, to find the atomic weight of the metal, we divide the mass of the metal by the number of moles of the metal:
Atomic weight of the metal = Mass of the metal / Number of moles of the metal = 111g / 6.24 moles ≈ 17.79 g/mol
Therefore, the atomic weight of the metal (assuming its valency to be 2) is approximately 17.79 g/mol.
M = metal
we have MCl2
1 mol chlorine and 1/2 mol of metal
35.5 g chlorine in a mol
111 - 35.5 = 75.5 g of M
so M is 75.5 * 2 = 151 g/mol
maybe Europium