Question

A $10$ per (w/w) solution of cane sugar has undergone partial inversion according to the reaction:

$Sucrose+Water\to Glucose+Fructose$

If the boiling point of solution is ${100.27}^{\circ }C$ :

(a) what is the average mass of the dissolved materials?

(b) what fraction of the sugar has inverted? $K_b\left(H_{2}O\right)=0.512Kmo{l}^{-1}kg$

• A. $\left(a\right)105.32,\left(b\right)31.17$%
• B. $\left(a\right)210.65,\left(b\right)68.2$%
• C. $\left(a\right)272.6,\left(b\right)74.92$%
• D. $\left(a\right)304.45,\left(b\right)82.03$%

Answer 1

The correct option is B $\left(a\right)210.65,\left(b\right)68.2$%

(a) $10\%$ w/w solution means $10$ kg of solute and $90$ kg of solvent.

$\Delta T_b={100.27}^{o}C-{100}^{o}C={0.27}^{o}C$

$M_2=\frac{K_b{W}_2}{\Delta T_b{W}_1}=\frac{0.512\times 10}{0.27\times 90}=0.21065kg/mol=210.65g/mol$

Hence, the average mass of the dissolved materials is $210.65$ g/mol.

(b) $Sucrose+Water\to Glucose+Fructose$

The molar masses of sucrose, glucose and fructose are 342 g/mol, 180 g/mol and 180 g/mol respectively.

Assume initially $1$ mole of sucrose is present and $x$ moles of sucrose undergo inversion to form $x$ moles of glucose and $x$ moles of fructose. $(1-x)$ moles of sucrose remain unreacted.

The average molar mass of the mixture is $\frac{342(1-x)+180x+180x}{1-x+x+x}=210.65$.

$\frac{342+18x}{1+x}=210.65$

$342+18x=210.65+210.65x$

$131.35=192.65x$

$x=0.682$

Hence, the fraction of sugar that has inverted is $\frac{0.682}{1}\times 100=68.2$.