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Standard XII

Chemistry

Elevation in Boiling Point

18 g of gluco...

Question

$18 g$ of glucose, $C_{6} H_{12} O_{6}$ (Molar Mass $= 180 g / m o l$ ) is dissolved in $1 k g$ of water in a sauce pan. At what temperature will this solution boil?
$[K_{b}$ for water $= 0.52 K k g m o l^{- 1}$ , boiling point of pure water $= 373.15 K$ ]

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Solution

We calculate the change of the boiling point if we will use the equation:

$Δ T = K_{b} \times m o l a l i t y$

$m o l a l i t y = \frac{m o l e s o f s o l u t e}{W e i g h t o f s o l v e n t (i n k g)}$ = $\frac{18}{180}$

Substituting the value we get,

$Δ T_{b} = 0.05$

So, the temperature at which solution will boil $T_{b} =$ $373.15 + 0.05$ $=$ $373.2 K$

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18 g

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18 g of glucose, C6H12O6 (Molar Mass =180g/mol) is dissolved in 1 kg of water in a 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]

We calculate the change of the boiling point if we will use the equation:ΔT=Kb×molalitymolality=moles of soluteWeight of solvent(in kg)=18180Substituting the value we get, ΔTb=0.05So, the temperature at which solution will boilTb=373.15+0.05=373.2K

$18 g$ of glucose, $C_{6} H_{12} O_{6}$ (Molar Mass $= 180 g / m o l$ ) is dissolved in $1 k g$ of water in a sauce pan. At what temperature will this solution boil?
$[K_{b}$ for water $= 0.52 K k g m o l^{- 1}$ , boiling point of pure water $= 373.15 K$ ]