Answer:
a. [LiBr] = 2.70 m
b. Xm for LiBr = 0.1
c. 81% by mass CH₃CN
Explanation:
Solvent → Acetonitrile (CH₃CN)
Solute → LiBr, lithium bromide
We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g
This mass of solute is contained in 1L of solution
1 L = 1000 mL → 1mL = 1cm³
We determine solution mass by density
Solution density = Solution mass / Solution volume
Solution density . Solution volume = solution mass
0.824 g/cm³ . 1000 cm³ = 824 g
Mass of solution = 824 g (solvent + solute)
Mass of solute = 156.3 g
Mass of solvent = 824 g - 156.3 g = 667.7 g
Molality → Moles of solute in 1kg of solvent
We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg
Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality
Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)
Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles
Total moles = 16.3 + 1.8 = 18.1
Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1
Mass percentage → (Mass of solvent, <u>in this case</u> / Total mass) . 100
<u>We were asked for the acetonitrile</u> → (667.7 g / 824 g) . 100 = 81%
