Question

Elevation of boiling point of $1$ molar aqueous glucose solution (density$= \,1.2\,g\,m{l^{ - 1}}$ ) is
A) ${K_b}$
B) $1.20\,{K_b}$
C) $1.02\,{K_b}$
D) $0.98\,{K_b}$

Answer 1

It is a question of solutions where on adding a nonvolatile solute in solvent the vapor pressure of solution decreases and shows decrease as compared to the vapor pressure of pure solvent. Thus many properties also get affected by this, like as suggested in the above question that elevation of boiling point is seen. Use the formula of elevation of boiling point.

Complete step-by-step answer:
We have given that there is an elevation of boiling point of one molar aqueous solution of glucose, it means when glucose was added in the solvent which is water here, it acts as a nonvolatile solute and decreases the vapor pressure of solution. Now, we have given that there is a $1$ molar solution present it means one mole of glucose is mixed in $1000\,c{m^3}$ of solution.
Formula for elevation of boiling point can be written as- $\Delta {T_b} = \,{K_b} \times \dfrac{{{n_{solute}}}}{{mass\,of\,solvent\,(in\,kg)}}$
$1\,M\, = \,1\,mole$ of glucose in $1000\,c{m^3}$ of solution.
We can easily find out the mass of glucose by multiplying its molar mass with the number of moles, if you remember this formula.
$No.\,\,of\,mole = \,\dfrac{{given\,mass}}{{molar\,mass}}$
$given\,mass = \,moles\, \times molar\,mass$
On putting the values we get mass of glucose as, $mass\, = \,1 \times \,(180.56)\,g\,mo{l^{ - 1}}$ where $(180.56)\,g\,mo{l^{ - 1}}$ is the molar mass of glucose.
$Density\,of\,solution\, = \,\dfrac{{mass\,of\,solution}}{{volume\,of\,solution}}$
For finding mass of solution, we have to multiply density of the aqueous solution with volume of solution, as we have made solution in $1000\,c{m^3}$ of volume thus, we can write the mass as,
$mass\,of\,solution = \,density \times \,volume\, = \,1.2\, \times 1000\, = \,1200\,g$

Also, mass of solvent can be found by subtracting the mass of solute from mass of solution, thus we will get mass of solvent which we use further in the formula of elevation in boiling point.
$Mass\,of\,water = \,1200 - \,180.56 = \,1019.44\,g = \,1.01944\,Kg$
Putting values in the above formula of elevation of boiling point and we will get something like this on solving further we will get a value of $0.98\,{K_b}$ .
$\Delta {T_b} = \,{K_b} \times \dfrac{{1\,mol}}{{1.01944\,kg}}$ = $0.98\,{K_b}$

Hence the correct answer is option ‘D’.

Note: A very common mistake that mostly students do is that they put mass of solution in place of mass of solvent in the main formula of elevation of boiling point $\Delta {T_b} = \,{K_b} \times \dfrac{{{n_{solute}}}}{{mass\,of\,solvent\,(in\,kg)}}$ . Also we have to change the mass from grams to kilograms because in the expression only a kilogram unit is needed. Lastly, we have not given the value of ${K_b}$ hence we leave it as per but in many questions the value of ${K_b}$ is given sometimes.