(a) 18 g of glucose (molar mass = 180 g mol⁻¹) is dissolved in 1000 g of water in sauce pan. At what temperature will this solution boil ? (Kb for water = 0.52 K Kg mol⁻¹, boiling point of pure water = 373.15 K)
(b) State and explain briefly Henry's law.

Question

Accepted Answer

>

> [!TLDR]
> (a) Elevation in boiling point $\Delta T_b = K_b m$ gives a boiling point of 373.202 K. (b) Henry's law relates a gas's partial pressure to its mole fraction dissolved in a liquid.
>
> **(a)** Moles of glucose $= \dfrac{18}{180} = 0.1$ mol. Mass of water $= 1000\ 	ext{g} = 1\ 	ext{kg}$, so molality:
> $$m = \dfrac{0.1\ 	ext{mol}}{1\ 	ext{kg}} = 0.1\ 	ext{mol kg}^{-1}$$
>
> Elevation in boiling point:
> $$\Delta T_b = K_b 	imes m = 0.52 	imes 0.1 = 0.052\ 	ext{K}$$
>
> Boiling point of solution:
> $$T_b = 373.15 + 0.052 = 373.202\ 	ext{K}\ (\approx 100.05^{\circ}	ext{C})$$
>
> **(b) Henry's law**: At a constant temperature, the solubility of a gas in a liquid (expressed as its mole fraction in solution) is directly proportional to the partial pressure of that gas above the liquid surface:
> $$p = K_H\, x$$ …

Q.(a) 18 g of glucose (molar mass = 180 g mol⁻¹) is dissolved in 1000 g of water in sauce pan. At what temperature will this solution boil ? (Kb for water = 0.52 K Kg mol⁻¹, boiling point of pure water = 373.15 K)

Concept understanding — Elevation of Boiling Point

Why does salt make water boil hotter?

The precise statement

Why proportional to molality, not molarity?

A concrete example

Q.(a) 18 g of glucose (molar mass = 180 g mol⁻¹) is dissolved in 1000 g of water in sauce pan. At what temperature will this solution boil ? (Kb for water = 0.52 K Kg mol⁻¹, boiling point of pure water = 373.15 K)

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