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{tex}\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}{/tex}
Oxidation of one mole of water will require 2×96500=1.93×105C.
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The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d-orbital.
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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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Ankit Kumar Kumar 2 years, 5 months ago
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