A 74.28-g sample of ba(oh)2 is dissolved in enough water to make 2.450 liters of solution. how many ml of this solution must be
diluted with water in order to make 1.000 l of 0.100 m ba(oh)2?
1 answer:
First let us calculate the initial molarity of the 2.45 L
of solution. Molar mass = 171.34 g/mol
<span>moles Ba(OH)2 = 74.28 g * (1 mole / 171.34 g) = 0.4335 moles</span>
Molarity (M1) = 0.4335 moles / 2.45 L = 0.177 M
Now using the formula M1V1 = M2V2, we can calculate how
much to dilute (V1):
0.177 * V1 = 0.1 * 1
V1 = 0.56 L
<span>Therefore 0.56 L of the initial solution must be diluted
to 1 L to make 0.1 M</span>
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