1 g mixture, consisting of equal number of moles of carbonates of two alkali metals, required 44.26 ml of 0.5 M HCl for complete reaction. If the atomic mass of one metal is 7 . If mass of metal carbo

Question

1 g mixture, consisting of equal number of moles of carbonates of two alkali metals, required 44.26 ml of 0.5 M HCl for complete reaction. If the atomic mass of one metal is 7 . If mass of metal carbonate having metal of atomic mass 7 in the original mixture is X × 1 0 − 4 gram. Find the value of X .

TestHub · Accepted Answer

Question Explanation: Find the mass of a specific metal carbonate in a mixture.

Concept: Titration of Carbonate Mixtures.

Solution: An equimolar mixture of two alkali metal carbonates reacts with HCl . Since the moles of both carbonates are equal, and each mole requires 2 moles of HCl , we use the titration data to find the moles and then the specific mass of the carbonate containing the metal with atomic mass 7 .

Mass of metal A (atomic mass $= 7$ ) $= x$

Mass of metal $B = 1 - x$

$\frac{x}{2 \times 7 + 60} = \frac{1 - x}{2 M _{B} + 60}$

$\frac{x}{2 \times 7 + 60} \times 2 + \frac{1 - x}{2 M _{B} + 60} \times 2 = \frac{0.5 \times 44.26}{1000}$

$x = 0.4094$

Final answer: 4094

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