Chemistry help please...measuring solubility.

Question

how do i calculate the following amount in mol of solute in:

a mixture of 92 mL of 0.5 M HCI and 135 mL of 0.2M HCI

which of the following solutions is more concentrated? How can you tell the difference?

1.5 L of 0.015 M Mg(OH)2 or 500 mL of solution containing 1.75 g Mg(OH)2.

finally, what is the mass of the solute in the following solution?

200 mL of 1.5 M sodium hydroxide (NaOH)

Answer 1

Moles = molarity x Volume ( in L)

moles = 0.5 x 0.092 L = 0.046

moles = 0.2 x 0.135 L=0.0270

Moles Mg(OH)2 = 1.75 g / 58.33 g/mol = 0.0300

M = moles /L = 0.0300 / 0.500 = 0.0600 M

0.0600 > 0.0.15 so the second solution is more concentrated

Moles = 1.5 x 0.200 L =0.300

Mass = mol x molar mass = 0.300 x 40 g/mol = 1.2 g of NaOH