Question

7. State the mass of the solute required to prepare 125 mL of 0.25 M Ba(NO3)2 solution.

Answer 1

The mass of the solute is 8.2 grams

Step-by-step explanation

Given that;

The volume of the solute is 125mL

The molarity of the solution is 0.25M

Follow the steps below to find the mass of the solute

Note; the solute is Ba(NO3)2

Step 1; Convert the volume of the solution to L

Recall, that 1 mL is equivalent to 0.001L

Let v represent the volume of the solution in L

`\begin{gathered} \text{ 1mL }\rightarrow\text{ 0.001L} \\ \text{ 125mL }\rightarrow\text{ vL} \\ \text{ cross multiply} \\ \text{ 1mL}* vL\text{ }=\text{ 125mL }*\text{ 0.001L} \\ \text{ Isolate vL} \\ \text{ vL}=\text{ }\frac{125\cancel{mL}*0.001L}{1\cancel{mL}} \\ \text{ v }=\text{ 0.125L} \end{gathered}`

Hence, the volume in Liters is 0.125L

Step 2; Find the number of moles of the solute using the below formula

`\text{ n }=\text{ cv}`
`\begin{gathered} \text{ n }=\text{ 0.25 }*\text{ 0.125} \\ \text{ n }=\text{ 0.03125 mole} \end{gathered}`

STEP 3: Find the mass of the solute using the formula below

`\text{ Mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of Ba(NO3)2 is 261.337 g/mol

`\begin{gathered} \text{ 0.03125 }=(mass)/(261.337) \\ \text{ Cross multiply} \\ \text{ mass }=\text{ 0.03125 }*\text{ 261.337} \\ \text{ mass }=\text{ 8.2 grams} \end{gathered}`

Hence, the mass of the solute is 8.2 grams