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Elevation in Boiling Point

18 g of gluco...

Question

$18 g$ of glucose $(C_{6} H_{12} O_{6})$ dissolved in $1 k g$ of water is heated to boiling. The boiling point (in $K$ ) measured at $1 a t m$ pressure is closest to [Ebulioscopic constant, $K_{b}$ for water is $0.52 K k g m o l^{- 1}$ . Consider absolute zero to be $- {273.15}^{\circ} C]$ .

A

373.15

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B

373.10

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C

373.20

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D

373.25

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Solution

The correct option is C $373.20$
$△ T_{b} = K_{b} \times m$
where, $m = \frac{n u m b e r o f m o l e s o f s o l u t e}{m a s s o f s o l v e n t (K g)}$
$△ T_{b} = 0.52 \times \frac{\frac{18}{180}}{1}$

$△ T_{b} = 0.052$

$T_{b} - T_{b}^{o} = 0.052$

$T_{b} = 100 + 0.052 = 100 {.05}^{o} C = 373.2 K$

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Similar questions

18 g

(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}

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

18 g

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

= 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}

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

, boiling point of pure water

= 373.15 K

18 g

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}

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

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

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