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1 g mixture of equal number of mole of Li2CO3 and other metal carbonates (M2CO3) required 21.6 mL of 0.5N HCl for complete neutralisation reaction. What is the approximate atomic mass of the other metal?
Some Basic Concepts of Chemistry
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14/08/2021 10:55 am
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1 g mixture of equal number of mole of Li_{2}CO_{3} and other metal carbonates (M_{2}CO_{3}) required 21.6 mL of 0.5N HCl for complete neutralisation reaction. What is the approximate atomic mass of the other metal?
(a) 25
(b) 23
(c) 51
(d) 118
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14/08/2021 10:59 am
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Correct answer: (d) 118
Explanation:
Let x g of Li_{2}CO_{3} and (1  x) g of M_{2}CO_{3} present in given mixture
2 × Total moles of carbonates = moles of HCl and
2 x \((\frac{x}{74} + \frac{1  x}{2M + 60})\) = 21.6 x 0.5 x 10^{3}
x = 0.20
∵ \(\frac{x}{74}\) = \(\frac{1  x}{2M + 60}\)
M = 118
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