$18 g$ 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 = $373.15 K$ )

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Solution

Using the relation $Δ T_{b} = m K_{b}$

$m = \frac{\frac{18}{180}}{1} = 0.1$

Hence, $Δ T_{b} = 0.1 \times 0.52 = 0.052 K$

Hence, the boiling point of the solution will increase by $0.052 K$

New boiling point $= 373.15 + 0.052 = 373.202 k$

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Similar questions

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

]

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}

]

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

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}

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18 g of glucose, C6H12O6 (Molar mass = 180 g mol−1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil?(Kb for water = 0.52 K kg mol−1, boiling points of pure water = 373.15 K)

Using the relation ΔTb=mKbm=181801=0.1Hence, ΔTb=0.1×0.52=0.052 KHence, the boiling point of the solution will increase by 0.052 KNew boiling point =373.15+0.052=373.202 k

$18 g$ 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 = $373.15 K$ )

Using the relation $Δ T_{b} = m K_{b}$

$m = \frac{\frac{18}{180}}{1} = 0.1$

Hence, $Δ T_{b} = 0.1 \times 0.52 = 0.052 K$

Hence, the boiling point of the solution will increase by $0.052 K$

New boiling point $= 373.15 + 0.052 = 373.202 k$