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# 6gm of urea and 9gm of …

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6gm of urea and 9gm of glucose are dissolved in in 300 gm of water find the boiling point of solution

Preeti Dabral 1 month, 2 weeks ago

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,

{tex}\begin{aligned} & \text { number of moles of urea }=\frac{\text { given weight of urea }}{\text { molar mass of urea }} \\ & = \\ & \frac{6}{60}=0.1 \text { moles } \\ & \text { number of moles of glucose }=\frac{\text { given weight of glucose }}{\text { molar mass of glucose }} \\ & =\frac{9}{180}=0.05 \text { moles } \\ & \text { Total moles of solute }=0.1+0.05=0.15 \text { moles } \\ & \text { molality of solute }(b)=\frac{\text { moles of solute }}{\text { mass of solvent in } \mathrm{kg}} \\ & b=\frac{0.15 \times 1000}{300} \\ & b=0.5 \mathrm{M} \end{aligned}{/tex}

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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