Question

Given,$1.0{\text{ g}}$ of a mixture of equal number of moles of carbonates of two alkali metals required $44.4$ $mL$ of $0.50(N)$ $HCl$ for complete reaction. If the atomic weight of one of the metals is 7.0. Find the atomic weight of the other metal?

Answer 1

Alkali metals include lithium, sodium, potassium, rubidium, caesium and francium. Alkali metals constitute the group one in the s- block of the periodic table. These metals are called alkali metals because of their property of forming alkali i.e. water soluble bases when dissolved in water.

Complete answer:
Step 1: Assuming the formula of the carbonates of the alkali metal:
As alkali metals are monovalent i.e. metals with one as their valence. So, let the formula of the two carbonate of the alkali metal and their masses be:
${M_2}C{O_3} = a{\text{ grams}}$
And, ${M_2}^{'}C{O_3} = b{\text{ grams}}$
According to the question: $a + b = 1$
Step 2: Calculating the molar mass of the compounds:
According to the question, the atomic mass of one of the alkali metals is 7. So, the molar mass of the compound ${M_2}C{O_3} = (2 \times 7) + 12 + (3 \times 16) = 74{\text{ gram mo}}{{\text{l}}^{ - 1}}$
Similarly, assuming ‘$x$’ as the atomic mass of the other alkali metal. Then, the molar mass of the compound ${M^{'}}_2C{O_3} = (2 \times x) + 12 + (3 \times 16) = 2x + 60{\text{ gram mo}}{{\text{l}}^{ - 1}}$
Step 3: calculating the number of moles in each compound:
Numerically, Moles are defined as the ratio of given mass of the substance to the molecular mass of the substance.
According to the question, the 1 gram mixture contains an equal number of moles of the two carbonates of the alkali metals. So,
$\dfrac{a}{{74}} = \dfrac{b}{{2x + 60}}$
Step 4: Reaction between the mixtures of the carbonates of the alkali metal with $HCl$:
According to the question $44.4{\text{ mL}}$ of $0.50N{\text{ HCl}}$ is required for the reaction to complete. Therefore, total $\dfrac{{44.4}}{{1000}} \times 0.50 = 0.0222$ moles of $HCl$ is required for the reaction to complete.
$C{O^{2 - }}_3 + HCl \to 2MCl + C{O_2} + {H_2}O$
For the above reaction, two mole of $HCl$ reacts with one mole of $CO_3^{2 - }$
But, 0.0222 moles of $HCl$ will react with $\dfrac{{0.0222}}{2} = 0.0111$ moles of $CO_3^{2 - }$
Since, the number of moles of the two carbonates of the alkali metals is the same. Therefore,$\dfrac{{0.0111}}{2}$ moles of each metal carbonate is required for the reaction.
Step 5: Solving the equations:
$\dfrac{a}{{74}} = \dfrac{{0.0111}}{2}$
$a = 0.41$
Also, $b = 1 - a$
$b = 1 - 0.41$
$b = 0.59$
So, $\dfrac{b}{{2x + 60}} = \dfrac{{0.0111}}{2}$
$\Rightarrow \dfrac{{1 - a}}{{2x + 60}} = \dfrac{{0.0111}}{2}$
$\Rightarrow \dfrac{{1 - 0.41}}{{2x + 60}} = \dfrac{{0.0111}}{2}$
$x = 23.1$
Hence, the atomic weight of the other metal is $23.1$

Note:
Carbonates are salts of carbonic acid containing $CO_3^{2 - }$ ion. Carbonate is the simplest example of oxocarbon anion. Carbonate has a trigonal planar arrangement where one carbon atom is surrounded by three oxygen atoms. Carbonates are the conjugate base of bicarbonate or hydrogen carbonate ions.