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?

First let us calculate the initial molarity of the 2.45 L of solution. Molar mass = 171.34 g/mol

moles Ba (OH) 2 = 74.28 g * (1 mole / 171.34 g) = 0.4335 moles

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

Therefore 0.56 L of the initial solution must be diluted to 1 L to make 0.1 M