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C60

18 g of gluco...

Question

$18 g$ of glucose, $C_{6} H_{12} O_{6}$ , is dissolved in $1 k g$ of water in a saucepan.At what temperature will water boil at $1.013 bar$ ?
$K_{b} for water is 0.52 K k g m o l^{- 1}$

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Solution

Molality of glucose
Given,
Amount of glucose $= 18 g$

So,
Moles of glucose $= \frac{18 g}{180 g m o l^{- 1}} = 0.1 m o l$

Amount of solvent $= 1 k g$

Molality of glucose solution $= \frac{number of moles of solute}{amount of solute} = \frac{0.1}{1 k g}$

$\Rightarrow Molality of glucose solution = 0.1 m o l k g^{- 1}$

Temperature at which water will boil
We know, change in boiling point $Δ T_{b}$ is given by,
$Δ T_{b} = K_{b} \times m$
$= 0.52 K k g m o l^{- 1} \times 0.1 m o l k g^{- 1} = 0.052 K$

Since water boils at $373.15 K at 1.013 b a r$ pressure, therefore, the boiling point of solution will be $373.15 + 0.052 = 373.202 K$ .

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Similar questions

Q. 18 g of glucose,

C_{6} H_{12} O_{6}

, is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil (1.013 bar pressure)?

K_{b}

for water is 0.52 K kg

m o l^{- 1}

Q. Solve this:

Q . 18 g g l u \cos e d i s s o l v e i n 1 k g o f w a t e r i n a c o n t a i n e r a t w h a t t e m p e r a t u r e t h e w a t e r b o i l a t 1.013 b a r p r e s s u r e K b f o r w a t e r = 0.52 K K g m o l^{- 1}]

18 g

of glucose

(C_{6} H_{12} O_{6})

is dissolved in

1 k g

of water in a saucepan. At what temperature will the water boil at

1.013 b a r

[

K_{b}

for water is

0.52 K k g m o l^{- 1}

]

18 g

of glucose,

C_{6} H_{12} O_{6}

is dissolved in

1 k g

of water in a saucepan. At what temperature will the water boil
under atmospheric pressure?

K_{b}

for water is

0.52 K k g m o l^{- 1}

50g of glucose(180g) dissolved in 1kg of water .At what temperature the glucose solution will boils at 1.013 bar. Kb for water is 0.52 kgmol- (water bolis at 373.3 k at 1.013 bar).

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C60

Standard X Chemistry

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18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan.At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol−1

$18 g$ of glucose, $C_{6} H_{12} O_{6}$ , is dissolved in $1 k g$ of water in a saucepan.At what temperature will water boil at $1.013 bar$ ?
$K_{b} for water is 0.52 K k g m o l^{- 1}$