More chemistry help...it's the math ones?

Question

what is the molarity of an aqueious NaOH solution made with 5 kg of water and 3.6 mol NaOH (molar mass = 40.00)



How much methanol, CH3 OH (molar mass = 32.05 g/mol) is needed to make a .90 m solution in 250 g of water

Answer 1

Molarity is defined as moles per liter (volume) of solution; if you add moles to a kg of water, that will give you molality. 3.6 mole of NaOH will have a mass of 40*3.6 g = 144g. The volume of 5kg of water is 5L. If the amount of NaOH is so small compared to the volume of water, we can assume that the total volume of the solution does not change after adding NaOH, so that the total volume remains 5L. The molarity is then 3.6/5 moles/liter = 0.72M. This is not the exact way to prepare a molar solution. The solution will be an accurate 0.72 MOLAL, however.

You have the same problem in the second part, since 0.9M is defined for liters, not grams, but you can approximate by assuming 250g will be 250mL after mixing. 0.9M of methanol will have 0.9*32.05 g in one liter, or 0.25*0.9*32.05 g in 250 mL.

Answer 2

Assuming that the density of water is 1.00 and the NaOH volume has negilible solution volume impact then

Molarity = 3.6 mol / 5 litres of solution = 0.72 M

A 0.90 m CH3OH solution is equivalent to 0.90 m CH3OH per 1000 g of solvent. So

Mass CH3OH = 0.9 x 32.05 x 250 / 1000 = 7.21 g