Question

Elevation of boiling point of a $1\text{M}$ aqueous glucose solution (density $=1.2\text{g/ml}$) is:
($K_b$ is molal boiling point elevation constant)

• A. $K_b$
• B. $1.20K_b$
• C. $1.02K_b$
• D. $0.98K_b$

Answer 1

The correct option is D $0.98K_b$

Given,
Molarity of glucose solution $=1\text{M}$
Assuming $1\text{L}$ of the solution,
Volume of the solution $=1\text{L}$
Mass of the solution $=1200\text{g}$ ( given density $=1.2\text{g/ml}$)
Number of moles of glucose $=1\text{mol}$
Mass of glucose $=180\text{g}$
Relation between molality and molarity may be written as:
$\text{molality}=\frac{1000\times \text{molarity}}{\text{mass of the solution - mass of the solute}}$
Elevation in boiling point is given by
$△$$T_b=i{K}_{b}m$
where,
$△T_b$ is the elevation in boing point
$i$ is the van't Hoff factor
$K_b$ is the molal boiling point elevation constant
$m$ is the molality

$\therefore △T_b=1$$\times$$K_b$$\times$$\frac{1000}{(1200-180)}$
$\Rightarrow △$$T_b=0.98K_b$