[tex]\tt{\huge{\red{Solution:}}}[/tex]
Step 1: Calculate the number of moles of solute [Ba(OH)₂].
Note that the volume of solution must be expressed in liters (L) and one molar (M) solution is equal to one mole per liter (mol/L).
[tex]\begin{aligned} \text{moles of solute} & = \text{molarity} \times \text{volume of solution} \\ & = 0.06 \: M \times \text{2.50 L} \\ & = \text{0.06 mol/L} \times \text{2.50 L} \\ & = \text{0.15 mol} \end{aligned}[/tex]
Final Step: Calculate the mass of solute.
Note that the molar mass of Ba(OH)₂ is 171.3 g/mol.
[tex]\begin{aligned} \text{mass of solute} & = \text{moles of solute} \times \text{molar mass of solute} \\ & = \text{0.15 mol} \times \text{171.3 g/mol} \\ & = \boxed{\text{25.7 g}} \end{aligned}[/tex]
Hence, 25.7 g of Ba(OH)₂ is required to prepare 2.50 L of a 0.06 M solution.
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Note: Kindly swipe the screen to the left to see the continuation of the answer on the right side.
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