A solution of 10.m koh was used to prepare 2.00L of 0.500 M KoH how many millilitres of the original KOH solution are needed
I don't think so. I assume that 10 m is really 10 M. M stands for molarity; m stands for molality.
mLKOH x 10 M = 2000 mL x 0.500 M
mL KOH = 2000 x 0.500/10 = ?
150.ml
To find out how many milliliters of the original KOH solution are needed to prepare 2.00 L of 0.500 M KOH, we can use the formula:
M1 x V1 = M2 x V2
Where:
M1 = initial molarity of the KOH solution
V1 = volume (in liters) of the initial KOH solution
M2 = desired molarity of the final KOH solution
V2 = volume (in liters) of the final KOH solution
Let's plug in the values:
M1 = 10.0 M (given)
V1 = unknown (we need to find this)
M2 = 0.500 M (given)
V2 = 2.00 L (given)
Using the formula, we can rearrange it to solve for V1:
V1 = (M2 x V2) / M1
V1 = (0.500 M x 2.00 L) / 10.0 M
V1 = 0.100 L
To convert liters into milliliters, multiply by 1000:
V1 = 0.100 L x 1000 mL/L
V1 = 100 mL
Therefore, 100 milliliters of the original KOH solution are needed to prepare 2.00 L of 0.500 M KOH.
To find the number of millilitres of the original KOH solution needed, we need to use the equation for dilution:
M1V1 = M2V2
Where:
M1 = initial concentration of the solution
V1 = initial volume of the solution
M2 = final concentration of the solution
V2 = final volume of the solution
Given:
M1 = 10.0 M (concentration of the original KOH solution)
V1 = ? (unknown volume of the original KOH solution)
M2 = 0.500 M (final concentration of the solution)
V2 = 2.00 L (final volume of the solution)
Rearranging the equation and solving for V1, we get:
V1 = (M2 * V2) / M1
Substituting the given values into the equation:
V1 = (0.500 M * 2.00 L) / 10.0 M
V1 = 0.100 L
Since the volume is asked in millilitres, we need to convert liters to millilitres:
1 L = 1000 mL
Therefore, the answer is:
V1 = 0.100 L * 1000 mL/L
V1 = 100 mL
So, 100 millilitres of the original KOH solution are needed to prepare 2.00 liters of 0.500 M KOH.