Question

How would you prepare 250 ml of 0.150 M Mg(NO3)2(aq) ?

Answer 1

Molarity = number of moles of Mg(NO3)2 / volume of solution in L

0.150 M = number of moles of Mg(NO3)2 / 0.250

number of moles of Mg(NO3)2 = 0.150 * 0.250 = 0.0375 mole

Number of moles of Mg(NO3)2 = mass of Mg(NO3)2 / molar mass of Mg(NO3)2

0.0375 mole = mass of Mg(NO3)2 / 148.3 g/mol

Mass of Mg(NO3)2 = 0.0375 mole * 148.3 g/mol = 5.56 g

Therefore, the mass of Mg(NO3)2 = 5.56 g

Therefore, by taking 5.56 g of Mg(NO3)2 we can prepare 250 ml of 0.150 M Mg(NO3)2.