Question

How many grams of Mg(NO₃)₂ must be used to make 250ml of a 1.842M solution of Mg(NO₃)₂ ? What is the percent composition by mass of Mg(NO₃)₂ , the mole fraction of Mg(NO₃)₂, and the molality of Mg(NO₃)₂ of this solution? Assume the density of the solution is 1.0g/mol.

Answer 1

¡) Required molarity of Mg(NO3)2 = 1.842M = 1.842mol/L

Molar mass of Mg(NO3)2 = 148.32g/mol

mass of Mg(NO3)2 required for 1000ml = 148.32g/mol × 1.842mol = 273.205g

mass of Mg(NO3)2 required for 250ml = 273.205g/4 = 68.3013g

ii) Volume of water = 250ml

Mass of water = 250g

Mass of Mg(NO3)2 = 68.3013g

Mass of water + Mg(NO3)2 = 318.3013

Mass percent of Mg(NO3)2 =( 68.3013g/318.3013g )× 100 = 21.46%

iii) Percent of Mg(NO3)2 (w/v) =( 68.3013g/250ml)×100ml = 27.32%

iv) No of mole of water = 13.8773mol

No of mole of Mg(NO3)2 = 0.4605mole

Total mole = 14.3378

mole fraction of Mg(NO3)2 = 0.4605/14.3378 = 0.03212

v) No of mole of Mg(NO3)2 = 0.4605mol

Mass of Water = 0.250kg

Molality of Mg(NO3)2 = 0.4605mol/0.250kg = 1.842m