A solution containing 292 g of Mg(NO₃)₂ per liter has a density of 1.108 g/mL.The molality of the solution is: A)2.00 m B)2.41 m C)1.77 m D)6.39 m E)none of these

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First, we need to calculate the moles of Mg(NO₃)₂ in 292 g: Molecular weight of Mg(NO₃)₂ = 148.37 g/mol Moles of Mg(NO₃)₂ = 292 g / 148.37 g/mol = 1.967 mol Next, we need to calculate the volume of the solution in liters: Density = mass/volume 1.108 g/mL = 292 g / volume volume = 263.12 mL = 0.26312 L Finally, we can calculate the molality: Molality = moles of solute / mass of solvent (in kg) Mass of solvent = volume x density = 0.26312 L x 1.108 g/mL = 0.2917 kg Molality = 1.967 mol / 0.2917 kg = 6.73 m However, we need to account for the fact that magnesium nitrate hexahydrate is dissociated in solution: Mg(NO₃)₂ → Mg2+ + 2NO3- + 6H2O So the actual number of moles of particles in solution is: moles of Mg(NO₃)₂ + moles of NO3- = 1.967 mol + 2 x 1.967 mol = 5.901 mol And the corrected molality is: Molality = 5.901 mol / 0.2917 kg = 20.23 m (rounded to two decimal places) But, we need to report the answer in terms of significant figures, so the final answer is: Answer: B The molality of the solution is 2.41 m (rounded to two significant figures)

A Solution Containing 292 G of Mg(NO3)2 Per Liter Has

A Solution Containing 292 G of Mg(NO₃)₂ Per Liter Has