Question

The volume of $0 \cdot 5{\text{ M}}$ ${H_3}P{O_4}$ that completely dissolves $3 \cdot 1{\text{ g}}$ copper carbonate is?
(Molecular mass of copper carbonate$ = 124{\text{ g/mol}}$)
A. $55.5 mL$
B. $45.5mL$
C. $35.5mL$
D. $33.3 mL$

Answer 1

Hint: From the given data in question first find out the reaction scheme for dissolution reaction. One can get the ratio of the reactants from the reaction scheme. Mass and molecular weight values are given which means one can find out the number of moles used of each reactant in the given reaction. One can put these values accordingly by using the specific equation to find out the volume.

Complete step by step answer:
- First of all we will calculate the number of moles of copper carbonate by using the data given in the question itself.
\[Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}copper{\text{ }}carbonate = \dfrac{{{\text{Total mass}}}}{{{\text{Molecular mass}}}} = \dfrac{{3 \cdot 1}}{{124}} = 0 \cdot 025{\text{ moles}}\]
-As we got the value of the number of moles of copper carbonate we can calculate the number of moles of ${H_3}P{O_4}$ by drawing a balanced reaction scheme as follows,
$2{H_3}P{O_4} + 3CuC{O_3} \to C{u_3}{(P{O_4})_2} + 3C{O_2} + 3{H_2}O$
-From the above-balanced reaction we now know that when one mole of ${H_3}P{O_4}$ gets reacted
$1.5{\text{ mole}}$ of copper, carbonate gets reacted. So, for $0 \cdot 025{\text{ mole}}$ of copper carbonate, we need to calculate the number of moles for ${H_3}P{O_4}$ as follows,
Number of moles of ${H_3}P{O_4}$$ = \dfrac{{1 \times 0 \cdot 025}}{{1 \cdot 5}} = 0 \cdot 017{\text{ mole}}$
- Now by using the following formula we can know the value of volume,
${M_1}{V_1} = {M_2}{V_2}$
$0 \cdot 5 \times {V_1} = 0 \cdot 017 \times 1000$
${V_1} = \dfrac{{0 \cdot 017 \times 1000}}{{0 \cdot 5}} = 33 \cdot 3{\text{ mL}}$
Therefore, the volume needed for complete dissolution is $33.3mL$ which shows option D as a correct choice.

Note: For a solution where two solutes are completely dissolved in it then the equivalents of one solute are always equal to equivalents of the second solute present in the solution. To find the number of moles contribution of each solute present to the whole solution, the best method is to draw a balanced reaction scheme which will give the individual moles ratio of solutes present in the solution.