Chemistry, asked by syedismailart, 4 months ago

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

Answered by prannithelango
1

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.

Answered by soniatiwari214
0

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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