Question

6 gram of Urea and 9 gram of glucose are deserve in 300 gram of water find tha boiling point of solution​

Answer 1

Answer:

hope this is the answer

Answer 2

The boiling point of the solution is 273.41 k

GIVEN

Weight of urea = 6 grams

weight of Glucose = 9 grams.

Weight of water = 300 grams

TO FIND

The boiling point of the solution.

SoOLUTION

We can simply solve the given problem as follows-

To calculate the boiling point of the solution, we will apply the following formula -

∆Tₒ = Kb× b × i (eq 1)

Where,

∆Tₒ = boiling point elevation

Kb = ebullioscopic constant of the solvent ( kb = o.515 kg/mol)

b = molality of the solute

i = van't Hoff factor of solute ( since urea and glucose are non-ionic compounds, so, i= 1)

We know that,

Boiling point elevation is defined as :

ΔTₒ = T₁ - T ........(eq2)

where,

T₁ = boiling point of the solution

T = boiling point of the solvent (T = 273.15 k of water)

Now,

$number \: of \: moles \: of \: urea = \frac{given \: weight \: of \: urea}{molar \: mass \: of \: urea}$

=

$\frac{6}{60} = 0.1moles$

$number \: of \: moles \: of \: glucose = \frac{given \: weight \: of \: glucose}{molar \: mass \: of \: glucose}$

$= \frac{9}{180} = 0.05moles$

Total moles of solute = 0.1+0.05 = 0.15 moles

$molality \: of \: solute(b) = \frac{moles \: of \: solute}{mass \: of \: solvent \: in \: kg \: }$

$b = \frac{0.15 \times 1000}{300}$

b = 0.5 M

putting the values in ( eq 1) we have

ΔTₒ = 0.512 × 0.5 × 1

ΔTₒ = 0.26 k

Now, putting the value of ΔTₒ in (eq 2)

ΔTₒ = T₁ - T

0.26 = T₁ - 273.15

T₁ = 273.15 + 0.26

T₁ = 273.41 k

Hence, The boiling point of the solution is 273.41 k

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