1.483 g of Mg(NO3)2 to make 100.0 mL of 0.1000 M solution.
21.163 g of Sr(NO3)2 to make 1000.0 mL of 0.10000 M solution.
First, determine the molar masses of the substances. Start by looking up the atomic weights of the involved elements.
Atomic weight magnesium = 24.305
Atomic weight nitrogen = 14.0067
Atomic weight oxygen = 15.999
Atomic weight strontium = 87.62
Molar mass Mg(NO3)3 = 24.305 + 2 * 14.0067 + 6 * 15.999 = 148.3124 g/mol
Molar mass Sr(NO3)2 = 87.62 + 2 * 14.0067 + 6 * 15.999 = 211.6274 g/mol
Now calculate how many moles of Mg(NO3)2 you need. Since molarity is defined as moles per liter, multiply the desired molarity by the desired volume. So
0.1000 M * 0.1000 liter = 0.01000 moles
Finally, multiple the number of moles by the molar mass, so
0.01000 mol * 148.3124 g/mol = 1.483124 g
Rounding to 4 significant figures gives 1.483 g
To make the Sr(NO3)2 solution, we've already calculated the molar mass of Sr(NO3)2, so we just need to calculate the number of moles needed. Since we want exactly 1 liter at 0.10000M, we need 0.10000 moles. So
0.10000 mol * 211.6274 g/mol = 21.16274 g
Rounding to 5 significant figures gives 21.163 g
