1 gm of a mixture of equal number of moles of Li 2 CO 3 and M 2 CO 3 required 44.44 ml of 0.5 M HCl for completion of the reaction : L i 2 C O 3 + 2 HCl → 2 LiCl + H 2 O + C O 2 M 2

Question

1 gm of a mixture of equal number of moles of $Li_{2} CO_{3}$ and $M_{2} CO_{3}$ required 44.44 ml of 0.5 M HCl for completion of the reaction :

$L i_{2} C O_{3} + 2 HCl \to 2 LiCl + H_{2} O + C O_{2}$

$M_{2} C O_{3} + 2 HCl \to 2 MCl + H_{2} O + C O_{2}$

If the atomic mass of Li is 7, and atomic mass of metal 'M' is X amu then nearest integer value of (M–16) is-

TestHub · Accepted Answer

Let each species be ' a ' moles & M be atomic mass of metal. $a \times (2 \times 7 + 12 + 48) + a (2 M + 12 + 48) = 1$ ....(i)

And a moles of each carbonate reacts with 2 a moles of HCl .

Hence $\to 4 a = (44.44 \times 0.5 \times 1 0^{- 3})$

$a = 11.11 \times 0.5 \times 1 0^{- 3} .. \dots (ii)$

From eq. (i) & (ii) $⟶ M = 23 gm / mol$

$M - 16 = 7$

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