Question

What volume ofwater is used to preparea 3.5 M solution if45 grams of Na2CO3is used?

Answer 1

We are asked to prepare a solution. They give us the molarity (3.5M) and grams of solute (45g of Na2CO3). Now, molarity is defined as moles of solute per liter of solution and has the following equation:

`Molarity=(MolesSolute)/(Lsolution)`

We must then find the moles equivalent to 45 grams of Na2CO3, for that we divide the grams by the molar mass of the compound. The molar mass of Na2CO3 is 105.99g/mol. So, the moles of NaCO will be:

`molNa_2CO_3=45gNa_2CO_3*(1molNa_2CO_3)/(105.99gNa_2CO_3)=0.42molNa_2CO_3`

Now, we solve the volume of the molarity equation. The volume of the solution has to be:

`\begin{gathered} Volume=(MolesSolute)/(Molarity) \\ V=(0.42mol)/(3.5M)=0.12L=120mL \end{gathered}`

The volume used to prepare the solution is 120mL

Answer: Third option. 120mL