Question

Elevation of boiling point of 1 molar glucose solution \[\left( {density = 1.2g/ml} \right)\] is:
 $
  (a){K_b} \\
  (b)1.20{K_b} \\
  (c)1.02{K_b} \\
  (d)0.98{K_b} \\
 $

Answer 1

Hint: Calculate the Van’t Hoff factor of glucose. The ratio of observed value of colligative property to its calculated value is called Van’t Hoff factor. The Van`t Hoff factor for 1 molal aqueous solution of glucose is 1.

Complete step by step answer:
We are given 1 molal glucose solution. This means 1 mole of glucose is dissolved in $1000c{m^3}$ of solution.
Molar mass of Glucose \[ = 180.56g/mol\]
Mass of Glucose= \[1 \times 180.56 = 180.56g\]
Mass of the solution = $density \times volume = 1.2 \times 1000 = 1200g$
Mass of water $ = 1200 - 108.56 = 1019.44g$
$1019.44g = 1.019Kg$
Let
${T_b}$ = elevation in boiling point
$i$ = van't hoff factor
$m$ = molality
${K_b}$ = molal boiling point constant.
The formula for elevation in boiling point is:
$
  \Delta {T_b} = im{K_b} \\
  \Delta {T_b} = i{K_b}\dfrac{n}{{mas{\text{s of solvent}}}} \\
  \Delta {T_b} = \left( 1 \right)\left( {{K_b}} \right)(\dfrac{{1mol}}{{1.099kg}}) \\
  \Delta {T_b} = 0.98{K_b} \\
 $

So, the correct answer is Option D.

Note:
It is a fact that the boiling point of a solvent will be higher when another compound is added i.e. the boiling point of a pure solvent is less than that of solution. This phenomena is known as the elevation in boiling point.