Question

50.00 mL of a solution containing 0.15 M CH2 (CO2 H)2 and 0.020 M MnSO4

Answer 1

Answer:

$m_{solute}=0.78gCH_2 (CO_2 H)_2$

Explanation:

Hello.

In this case, for the given volume and molarity of malonic acid (solute), we can compute the mass by first computing the moles considering the definition of molarity:

$M=\frac{n_{solute}}{V_{solution}} \\\\n_{solute}=M*V_{solution}=0.15mol/L*50.00mL*\frac{1L}{1000mL}\\ \\n_{solute}=0.0075molCH_2 (CO_2 H)_2$

Then, since the molar mass of the malonic acid is 104.1 g/mol, we can compute the required mass in g of malonic acid for the preparation of that solution as shown below:

$m_{solute}=0.0075molCH_2 (CO_2 H)_2*\frac{104.1gCH_2 (CO_2 H)_2}{1molCH_2 (CO_2 H)_2}\\ \\m_{solute}=0.78gCH_2 (CO_2 H)_2$

Notice that the shortest number is the molarity (0.15 M) with two significant figures, that is why the answer is displayed with two significant figures too.

Regards.