What is the molarity of 550.0 mL of KOH solution, that reacts completely with 675 mL of a 0.450 M CrCl3 solution?
balance the equation:
3KOH+ CrCl3>>3KCl + Cr(OH)3
notice for each mole of CrCl3, you use three moles of KOH.
moles of CrCl3=.450*.675
moles of KOH=3*.450*.675
molarity of KOH= moles KOH/.550
You made this so easy! Thank you!
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To calculate the molarity of the KOH solution, we need to use the equation:
M1V1 = M2V2
Where:
M1 = Molarity of the first solution (CrCl3 solution)
V1 = Volume of the first solution (675 mL)
M2 = Molarity of the second solution (KOH solution)
V2 = Volume of the second solution (550.0 mL)
Now let's plug in the given values into the equation and solve for M2:
(0.450 M)(675 mL) = M2(550.0 mL)
To solve for M2, divide both sides of the equation by 550.0 mL:
M2 = (0.450 M)(675 mL) / 550.0 mL
M2 = 0.55 M
Therefore, the molarity of the KOH solution is 0.55 M.