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suppose, V ml is the volume required from the stock solution,
and then the number of moles in that V will be, n1
n1 = 10 mol/dm3 x (v/1000)
n1 = v/100
This amount of moles are in 250mL of 0.375M solution,
0.375 = (v/100)/250 x 1000
0.375 = v/25
v = 9.375 mL
But the amount you can measure by a measuring cylinder (10 mL) is 9.4 mL. (0.1 mL accuracy)
Therefore the answer is 9.4 mL
Assuming that xL is required.
the answer is 0.009375L=9.375mL.
And you need a graduated container, such as a cylinder
It can contain 10mL.
but the smallest scale is 0.1mL.
9.375mL in it and you can only read 9.4 ml.
it is the container that makes difference.
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