What volume of 2.5% (m/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?

Chemistry

1 answer:

TiliK225 [7]3 years ago

6 0

Hey there!:

According to lew of dilution :

M*V = M'*V'

Where :

M = concentration of stock solution = 5.0% ( m / V )

V = volume of stock solution = 125 mL

M' = concentration of dilute solution = 2.5% ( m/V)

V' = volume of dilute solution = ??

Plug the values we get:

V' = m*V / M'

V' = 5.0 * 125 / 2.5

V' = 625 / 2.5

V' = 250 mL

Therefore , the volume prepared is 250 mL

Hope that helps!

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