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        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

							One mole of metal carbonate, on heating, decomposes to form 1 mole of oxide with the evolution of 1 mole of CO2 gas.MCO3 MO + CO2 ↑1 mole  1 mole 1 moleThus 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 gConversion of 0.44 g of CO2 into mole of CO244 of CO2= 1 mole of CO2∴ 0.44 g of CO2 = 0.44/44 = 0.01 moleFrom the equation, it is also obvious that1 mole of CO2 ≡ 1 mole of MCO3=∴ 0.01 mole of CO2 ≡ 0.01 mole of MCO3Thus the given mixture has 0.01 mole of MCO3 which will yield 0.01 mole of MO.From the problem, we also know that16 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 mlTotal volume of N HCI added = 100 ml∴ Volume of N HCI used = 100 – 40 = 60 mlHence 60 ml of N HCI is used in neutralizing BaO and MO according to the following reaction.MO + 2HCI ↓ MCI2 + H2OThus 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 mlEq. wt. of BaO = 138 +16/2 = 7740 ml of N HCI = 40 ml of N BaO = 40 *77/1000 = 3.08 g of BaOTotal wt. of the oxides = 3.64 g∴ Wt. of MO in the mixture = 3.64 – 3.08 = 0.56 gBut we know that 0.01 mole of MO is present in the residue it means0.01 mole of MO = 0.56 g of MO1 mole of MO = 56 g of MOSuppose the atomic weight of M = aThen the mol. wt. of MO = a + 16We 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 = 40Hence, the metal M must be calciumALTERNATIVE 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 moleNumber of moles of MCO3 ≡ 0.01 mole[∵ MCO3 MO + CO2]Volume of 1N HCI in which residue is dissolved = 100 mlVolume of 1N HCI used for dissolution = (100 – 2.5 * 16) ml = 60 ml= 60/1000 = 0.06 equivalentsThe chemical equation for dissolution can be written as + 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 = 0.03Number of moles of BaO = (0.03 – 0.01) = 0.02 molesMolecular weight of BaO = 138 + 16 = 154∴ Weight of BaO = (0.02 * 154) g = 30.8 gWeight of MCO3 = (4.08 – 3.08) = 1.0 gSince weight of 0.01 mole of MCO3 = 1.0 g∴ Mol. wt. of MCO3 = 1/0.01 = 100Hence atomic weight of unknown M = (100 - 60) = 40The atomic weight of metal is 40 so the metal M is Ca

5 years ago
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