The mole fraction of solute in a 3.87 m aqueous solution is 0.0697
<h3>
calculation</h3>
molality = moles of the solute/Kg of the solvent
3.87 m dissolve in 1 Kg of water= 1000g
find the moles of water= mass/molar mass
that is 1000 g/ 18 g/mol= 55.56 moles
mole of solute = 3.87 moles
mole fraction is = moles of solute/moles of solvent
that is 3.87/ 55.56 = 0.0697
We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
Answer:
it's lead (ii) nitrate the name
