Respuesta :
Given:
Mass of benzene (solvent) = 100.0g = 0.1 kg
Molar mass of solid (solute) = 89 g/mol
Elevation in b.pt (ΔTb) = 2.42 C
B.pt elevation constant k = 2.53 C.kg/mol
To determine:
The mass of the solid to be added
Explanation:
The elevation in b,pt is proportional to the molality (m) of the solute ;
ΔTb = k.m
where m = molality = moles of solute/kg of solvent
In this case we have:
ΔTb = k(benzene) * moles of solid/kg of benzene
2.42 = 2.53 * moles of solid/0.1
moles of solid = 0.0956 moles
The molar mass of solid = 89 g/mol
Thus, mass of solid = 0.0956 moles * 89 g/mol = 8.508 g
Ans: Mass of solid to be added is 8.508 g
Here we have to get the amount of solid required to raise the temperature of 100 g of benzene 2.42°C.
To raise the temperature of 100 g of benzene B) 8.52 g of the solid is needed.
We know, Δ[tex]T_{b}[/tex] = [tex]K_{b}[/tex]×[tex]C_{m}[/tex].
Where Δ[tex]T_{b}[/tex] is the elevation of the boiling point i.e. 2.42°C.
[tex]K_{b}[/tex] is the boiling point elevation constant of benzene i.e. 2.53°C.kg/mol.
[tex]C_{m}[/tex] is the molality of the the solute i.e. the solid.
Now the [tex]C_{m}[/tex] can be obtained by plugging the values in the equation.
2.42°C = 2.53°C.kg/mol × [tex]C_{m}[/tex]
Or, [tex]C_{m}[/tex] = [tex]\frac{2.42}{2.53}[/tex] mol/kg
Or, [tex]C_{m}[/tex] = 0.956 mol/kg.
Now 1 mole of the solid is equivalent to 89.0 g.
Thus 0.956 mol is equivalent to 0.956×89 = 85.13 g.
If 1000 g or 1 Kg of benzene contains 85.13 g of the solid its molality will be 0.956.
Thus in 100 g or 0.10 kg benzene will contain 85.13×0.10 = 8.51 g of the solid.
Thus 8.51 g of the solid is needed to raise the temperature.
