Question

For $90$ gm glucose and $120$ gm urea dissolved in a $1.46$ kg aqueous solution, then what will be the boiling point of the solution at $1$ bar pressure? (\[{k_b} = 0.512\dfrac{{^ \circ C.kg}}{{mol}}\], , molecular weight of glucose and urea are \[180\] and \[60\dfrac{g}{{mol}}\] respectively)
A.${100.876^ \circ }C$
B.\[{101.024^ \circ }C\,\]
C.\[{100.248^ \circ }C\]
D.${100.007^ \circ }C$

Answer 1

Hint: In order to solve this question first of all we will calculate the number of moles of all the solutes, add them and calculate the molality of the solution. On multiplying the molality with the boiling molal constant and get the change in temperature or the elevation of boiling point.

Complete answer:
In the above given question, we are given the mass of glucose and urea and required to calculate the change in boiling point of the aqueous solution.
Now first we need to calculate the moles of glucose and urea and the molecular weight if both of these compounds are already given in the question
Moles of glucose would be equal to the given mass with molecular weight which is
$moles = \dfrac{{given\,mass}}{{molecular\,mass}}$
$ \Rightarrow moles = \dfrac{{90}}{{180}}$
$ \Rightarrow moles = 0.5$
Moles of urea would be equal to the given mass of urea by its molecular weight
$moles = \dfrac{{given\,mass}}{{molecular\,mass}}$
$ \Rightarrow moles = \dfrac{{120}}{{60}}$
$ \Rightarrow moles = 2$
Now the total moles of solute in the aqueous solution are equal to $(0.5 + 2 = )2.5$ moles
Let’s calculate the molarity of the solute
Molality of a compound is defined as the total number of moles of the solute by the total weight of the solvent
Molality of the solutes in a $1.46$ kg aqueous solution is
$molality = \dfrac{{moles\,of\,solute}}{{weight\,of\,solvent}}$
$ \Rightarrow molality = \dfrac{{2.5}}{{1.46}}$
$ \Rightarrow molality = 1.712$
Now let’s calculate the elevation in boiling point which is equal to the product of molal boiling elevation constant to the molality of the solute, which is equal to
$\Delta {T_b} = {k_b} \times molality$
$ \Rightarrow \Delta {T_b} = 0.512 \times 1.712$
$ \Rightarrow \Delta {T_b} = 0.876$
The normal boiling point of any aqueous solution without any solute is ${100^ \circ }C$, if the elevation is boiling point is by ${0.876^ \circ }C$, then the new boiling point is ${100.876^ \circ }C$.
If $90$ gm glucose and $120$ gm urea dissolved in a $1.46$ kg aqueous solution, then the boiling point of the solution at $1$ bar pressure is ${100.876^ \circ }C$.

Note:
The addition of solute in aqueous solution to increase the boiling point or decrease the freezing point is used in our daily works including when the snow covers the roads then salt is sprinkled on it which lowers its boiling point thus making the ice melt faster.