How any grams of solid mg(no3)2 to dissolve in water to make your 0.1000 m solutions. specifically, write out the calculation to prepare 100.0ml of 0.1000m mg(no3)2 and 1000.0 ml of 0.10000m sr(no3)2

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

The density would be 1.622