10 grams of a non volatile and non dissociating solute is dissolved in 200 g of Benzene the resulting solutions boils at a temprature 81.2°c. Find the Molals mars of solute (Obullioscopic constant 2.53%molals).
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Answers
Δp=p
∘
x
2
0.50=100×
n
1
n
2
=100×
m×W
w×M
Here,
w= mass of solute = 10g
W= mass of solvent = 180g
M= molar mass of solvent =18g/mol
m= molar mass of solute =?
Substituting the value we get, m=200 g/mol
ΔT
b
=k
b
m=K
b
W
w
×
m
1000
=0.52×
180
10
×
200
1000
=0.14≈0.15
T
b
=100.15
∘
C.
Concept:
According to elevation in boiling point,
ΔTb = Kb n
where ΔTb is an elevation in boiling point, Kb is the Obullioscopic constant and m is the molality.
Given:
Mass of the solute, m = 10gm
Mass of solvent (Benzene) = 200 gm = 0.2 kg
The boiling point of the solution = 81.2°C
Obullioscopic constant = 2.53°C/mol
Find:
The molar mass of the solute.
Solution:
The initial boiling point of benzene = 80.1 °C
Elevation of boiling point (ΔTb) = (81.2-80-1) = 1.1 °C.
According to elevation in boiling point,
ΔTb = Kb m
ΔTb = Kb × moles of solute/Mass of solvent
ΔTb = Kb × (mass of solute/Molar mass)/Mass of solvent
The molar mass of solute,
M = (Kb × moles of solute)/(ΔTb ×Mass of solvent)
M = (2.53 × 10)/(1.1 × 0.2)
M = 115 g/mol
Hence, the molar mass of the solute is 115 gm/mol
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