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Grade: 10
        
4.08 g of a mixture of BaO and an unknown carbonate MCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 ml of 1 N HCI. The excess acid required 16 ml of 2.5 N NaOH solution for complete neutralization. Identify the metal M.
(At. wt. H = 1, C = 12, O = 16, CI = 35.5, Ba = 138)
5 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
							
One mole of metal carbonate, on heating, decomposes to form 1 mole of oxide with the evolution of 1 mole of CO2 gas.
MCO3 \overset{heat}{\rightarrow}MO + CO2
1 mole 1 mole 1 mole
Thus on heating, metal carbonate will loose weight corresponding to the weight of carbon dioxide.
From the given data, loss in wt. = 4.08 – 3.64 g = 0.44 g
Conversion of 0.44 g of CO2 into mole of CO2
44 of CO2= 1 mole of CO2
∴ 0.44 g of CO2 = 0.44/44 = 0.01 mole
From the equation, it is also obvious that
1 mole of CO2 ≡ 1 mole of MCO3=
∴ 0.01 mole of CO2 ≡ 0.01 mole of MCO3
Thus the given mixture has 0.01 mole of MCO3 which will yield 0.01 mole of MO.
From the problem, we also know that
16 ml of 2.5 N NaOH ≡ 16 ml of 2.5 n HCI [N1V1 = N2V2]
= 16 * 2.5 ml of N HCI = 40 ml of N HCI
∴ volume of N HCI remain unused = 40 ml
Total volume of N HCI added = 100 ml
∴ Volume of N HCI used = 100 – 40 = 60 ml
Hence 60 ml of N HCI is used in neutralizing BaO and MO according to the following reaction.
MO + 2HCI ↓ MCI2 + H2O
Thus 0.01 mole of MO ≡ 0.02 mole of HCI
= 0.02 * 1000 ml of N HCI
= 20 ml of N HCI
∴ Vol. of N HCI used = 60 – 20 = 40 ml
Eq. wt. of BaO = 138 +16/2 = 77
40 ml of N HCI = 40 ml of N BaO = 40 *77/1000 = 3.08 g of BaO
Total wt. of the oxides = 3.64 g
∴ Wt. of MO in the mixture = 3.64 – 3.08 = 0.56 g
But we know that 0.01 mole of MO is present in the residue it means
0.01 mole of MO = 0.56 g of MO
1 mole of MO = 56 g of MO
Suppose the atomic weight of M = a
Then the mol. wt. of MO = a + 16
We also know that 1 mole of MO (i.e. Mol. wt. of Mo) = 56
∴ a + 16 = 56 or a = 56 – 16 = 40
∴ The atomic weight of metal M = 40
Hence, the metal M must be calcium
ALTERNATIVE SOLUTION :
Weight of MCO3 and BaO = 4.08 g (given)
Weight of residue = 3.64 g (given)
∴ Weight of CO2 evolved on hating = (4.08 – 3.64) g= 0.44 g
= 0.44/44 = 0.01 mole
Number of moles of MCO3 ≡ 0.01 mole
[∵ MCO3 \overset{heat}{\rightarrow}MO + CO2]
Volume of 1N HCI in which residue is dissolved = 100 ml
Volume of 1N HCI used for dissolution = (100 – 2.5 * 16) ml = 60 ml
= 60/1000 = 0.06 equivalents
The chemical equation for dissolution can be written as
\underbrace{BaO + MO} + 4HCI → BaCI2 + MCI2 + 2H2O
Residue
[Number of moles of BaO and MO = 1 + 1 = 2]
Number of moles of BaO + Numebr of moles of MO = 0.06/2 = 0.03
Number of moles of BaO = (0.03 – 0.01)
= 0.02 moles
Molecular weight of BaO = 138 + 16 = 154
∴ Weight of BaO = (0.02 * 154) g
= 30.8 g
Weight of MCO3 = (4.08 – 3.08) = 1.0 g
Since weight of 0.01 mole of MCO3 = 1.0 g
∴ Mol. wt. of MCO3 = 1/0.01 = 100
Hence atomic weight of unknown M = (100 - 60) = 40
The atomic weight of metal is 40 so the metal M is Ca
5 years ago
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