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Standard XII

Chemistry

Elevation in Boiling Point

18 g of gluco...

Question

18 g of glucose, $C_{6} H_{12} O_{6}$ , 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 $m o l^{- 1}$ .

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Solution

$△ T = K_{b} \times \frac{w_{B} \times 1000}{m_{B} \times w_{A}}$
$△ T = 0.52 \times \frac{18 \times 1000}{180 \times 1000}$
$T - T_{0} = 0.052$
$T - 373 = 0.052$
$T = 373.052 K$

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Similar questions

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}

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}

(Molar mass = 180 g

m o l^{- 1})

is dissolved in

1 k g

of water in a saucepan. At what temperature will this solution boil?
(

K_{b}

for water =

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

, boiling points of pure water =

373.15 K

)

18g 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 (at 1 atm)? $K_{b}$ for water is $0.52 K k g m o l^{- 1}$ .

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

△T=Kb×wB×1000mB×wA△T=0.52×18×1000180×1000T−T0=0.052T−373=0.052T=373.052K

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 $m o l^{- 1}$ .

$△ T = K_{b} \times \frac{w_{B} \times 1000}{m_{B} \times w_{A}}$
$△ T = 0.52 \times \frac{18 \times 1000}{180 \times 1000}$
$T - T_{0} = 0.052$
$T - 373 = 0.052$
$T = 373.052 K$