Question

How many milliliters of 10.5 M HCl(aq) are needed to prepare 740.0 mL of 1.00 M HCl(aq)?

Answer 1

1)

M1 * V1 = M2 * V2

10.5 * V1 = 1 * 740

V1 = 70.48 ml .... Answer

2)

Initial Moles of solute = molarity * volume in liters = 1.2 * (61/1000) = 0.0732 moles

Now these moles are present in 238 ml of solution.

238 ml solution contains 0.0732 moles

So, 119 ml solution will contain = 0.0732 * 119 / 238 = 0.0366 moles of solute

Now total volume of solution = 119 + 113 = 232 ml = 0.232 liters

Final concentration = moles / volume in liters = 0.0366/0.232 = 0.158 M ....Answer

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