Question

The volume of $0.5M$ $H_{3}P{O}_4$ that completely dissolves $3.1$ g copper carbonate is (molecular mass of copper carbonate $=124$ g/mol) :

• A. $55.5$ mL
• B. $45.5$ mL
• C. $35.5$ mL
• D. $33.3$ mL

Answer 1

Answer: (D)

$33.3$ mL

As the molar mass of $CuC{O}_3$ is 124 g/mol.
So, number of moles in 3.1 g $=\frac{3.1}{124}=0.025$ moles

Now, the reaction is as follows:
$2H_{3}P{O}_4+3CuC{O}_3\to C{u}_3(P{O}_4{)}_2+3C{O}_2+3H_{2}O$

This shows that 1 mol of $H_{3}P{O}_4$ can dissolve 3/2 (1.5 ) mol of $CuC{O}_3$ completely.
So, now for 0.025 mol of $CuC{O}_3$ , the 0.017 mol of $H_{3}P{O}_4$ is required.

Now as we know.
$M_1{V}_1=M_2{V}_2$
$0.5\times V_1=60\times 1000$

$V_1=0.017\times \frac{1000}{0.5}=$ around 33.3 mL.