To determine how many moles of [tex]\( \text{Ba}(NO_3)_2 \)[/tex] are present in 0.25 liters of a 2.00 M [tex]\( \text{Ba}(NO_3)_2 \)[/tex] solution, we can use the formula for molarity:
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}}
\][/tex]
We are given:
- Volume of the solution, [tex]\( V = 0.25 \, \text{L} \)[/tex]
- Molarity of the solution, [tex]\( M = 2.00 \, \text{M} \)[/tex]
We need to find the moles of solute ([tex]\( \text{Ba}(NO_3)_2 \)[/tex]), which can be calculated by rearranging the formula to:
[tex]\[
\text{moles of solute} = \text{Molarity} \times \text{liters of solution}
\][/tex]
Substituting the given values:
[tex]\[
\text{moles of solute} = 2.00 \, \text{M} \times 0.25 \, \text{L}
\][/tex]
[tex]\[
\text{moles of solute} = 0.50 \, \text{mol}
\][/tex]
So, there are 0.50 moles of [tex]\( \text{Ba}(NO_3)_2 \)[/tex] in the 0.25 liters of the 2.00 M solution.
The correct answer is [tex]\( 0.50 \, \text{mol} \)[/tex].
