Answer:
[tex]\boxed {\boxed {\sf 0.123 \ mol \ Ba(OH)_2}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
The molarity of the solution is 0.600 M Ba(OH)₂. 1 molar (M) is also equal to 1 mole per liter, so the molarity is 0.600 moles of Ba(OH)₂ per liter.
The volume of the solution is 205 milliliters, however, we need to convert the volume to liters. Remember that 1 liter contains 1000 milliliters.
- [tex]\frac { 1 \ L}{1000 \ mL}[/tex]
- [tex]205 \ mL * \frac { 1 \ L}{1000 \ mL} = \frac {205}{1000} \ L = 0.205 \ L[/tex]
Now we know the molarity and volume, but the moles are still unknown.
- molarity = 0.600 mol Ba(OH)₂/ L
- moles of solute = x
- liters of solution = 0.205 L
Substitute these values into the formula.
[tex]0.600 \ mol \ Ba(OH)_2 /L = \frac{x}{0.205 \ L}[/tex]
We are solving for x or the moles of solute, so we must isolate the variable. It is being divided by 0.205 liters. The inverse of division is multiplication. Multiply both sides by 0.205 L.
[tex]0.205 \ L *0.600 \ mol \ Ba(OH)_2 /L = \frac{x}{0.205 \ L} * 0.205 \ L[/tex]
[tex]0.205 \ L *0.600 \ mol \ Ba(OH)_2 /L =x[/tex]
The units of liters cancel.
[tex]0.205 *0.600 \ mol \ Ba(OH)_2 =x[/tex]
[tex]0.123 \ mol \ Ba (OH)_2 = x[/tex]
The original measurements had at least 3 significant figures. Our answer currently has 3 sig figs, so we don't need to round.
There are 0.123 moles of barium hydroxide.
