A solution is prepared by adding 40.3 g of Mg(NO₃)₂ to 127 g of water. Calculate the mole fraction and molality of magnesium nitrate in this solution. A) Mole fraction = 0.0371; molality = 2.14 m B) Mole fraction = 0.0571; molality = 2.14 m C) Mole fraction = 0.0371; molality = 1.14 m D) Mole fraction = 0.0571; molality = 1.14 m E) None of the above

Question

Quizplus · Accepted Answer

Step 1: Calculate the moles of Mg(NO3)2.
m = 40.3 g
Molar mass = 148.3 g/mol
moles = m/M = 40.3/148.3 = 0.2714 mol

Step 2: Calculate the moles of water.
m = 127 g
Molar mass = 18.01 g/mol
moles = m/M = 127/18.01 = 7.053 mol

Step 3: Calculate the mole fraction of Mg(NO3)2
mole fraction = moles of Mg(NO3)2 / total moles
mole fraction = 0.2714 / (0.2714 + 7.053) = 0.0371

Step 4: Calculate molality
molality = moles of solute / mass of solvent (in kg)
mass of solvent = 127 g = 0.127 kg
molality = 0.2714 / 0.127 = 2.14 m

Therefore, the correct answer is A) Mole fraction = 0.0371; molality = 2.14 m.

A Solution Is Prepared by Adding 40