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1 mole glucose is added to 1 L of water ${K_{\mathbf{b}}}({H_2}O) = 0.512{\text{ K kg mol}}{{\text{e}}^{ - 1}}$boiling point of solution will be
(A) 373.512$^oC$
(B) 100.512$^oC$
(C) 99.488$^oC$
(D) 372.488$^oC$

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Last updated date: 25th Jun 2024
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Answer
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Hint: Molal elevation constant is defined as the elevation in boiling point of liquid produced when one mole of solute is dissolved in 1 kg i.e. 1000 g of the solvent.
- Addition of non-volatile solute here glucose in water, lowers the vapor pressure of water. Due to which the boiling point of water increases.

Complete Solution :
To answer the question why on addition of 1 mol glucose to 1 liter water the boiling point of water increases the following point need to be taken care of:
- The vapor pressure of the solvent is found to decrease in the presence of non-volatile solute such as glucose hence boiling point increases.
- Addition of non-volatile solute here glucose in water, lowers the vapor pressure of water. Due to which the boiling point of water increases.
- From the above discussion we come to know that the vapour pressure of liquid increases with increase of temperature. The liquid boils at the temperature at which its vapour pressure is equal to atmospheric pressure.
For example, water boils at 373.15 K because at this particular temperature the vapour pressure of water is 1.013 bar or (1 atmosphere).

- In a similar way like lowering of vapour pressure, the elevation of boiling point also depends on the number of solute molecules rather than the nature of the solute molecules.
$\Delta {T_b}$ = ${T_b}$-${T_b}^o$
- Where the $\Delta {T_b}$ is the elevation in the boiling point, ${T_b}$ be the boiling point of solution and ${T_b}^o$ be the boiling point of pure solvent.

Therefore,
${T_b}^o$ = 100$^oC$(BP of water)
$\Delta {T_b}$ = ${T_b}$-${T_b}^o$ = ${T_b}$-100$^oC$

- Experiments have shown that for a dilute solution $\Delta {T_b}$ (the elevation in the boiling point) is directly proportional to the Molal concentration of the solute in a solution given by the following equation:
$\Delta {T_b} = {K_{\mathbf{b}}}m$
${K_{\mathbf{b}}}$ is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant)
Here, m = Number of moles of solute per kg of solvent
- Here we are given 1 mole of the glucose is added to 1 L of water which is equivalent to the 1 kg of water as the density of water is one.
Therefore, m = $\dfrac{1}{1} = 1$

$\Delta {T_b} = {K_{\mathbf{b}}}m$
${T_b}$-100$^oC$ = $0.512{\text{ K kg mol}}{{\text{e}}^{ - 1}} \times 1{\text{ mole k}}{{\text{g}}^{ - 1}}$
${T_b}$ = 100$^oC$ + 0.512$^oC$ = 100.512$^oC$
So, the correct answer is “Option B”.

Note: One should be clear in the difference between Molal elevation constant and Molal depression constant:
Molal elevation constant is defined as the elevation in boiling point of liquid produced when one mole of solute is dissolved in 1 kg i.e. 1000 g of the solvent.

Molal depression constant:
- Molal depression constant is defined as the depression in boiling point of liquid produced when one mole of solute is dissolved in 1 kg i.e. 1000 g of the solvent.
- One should be very careful while using the equation else the wrong equation would yield the wrong answer.