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Standard XII

Chemistry

Osmotic Pressure

50g of glucos...

Question

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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Solution

$Δ T_{b} = K_{b} \times m w h e r e Δ T_{b} = e l e v a t i o n i n b o i l i n g p o i n t K_{b} = c o n s \tan t m = m o l a l i t y m o l a l i t y = n o . o f m o l e s i n 1 k g o f s o l v e n t m = \frac{\frac{m a s s g i v e n}{M o l a r m a s s}}{M a s s o f s o l v e n t (k g)} = \frac{\frac{50}{180}}{1} = 0.277 m Δ T_{b} = 0.52 \times 0.2777 = 0.1444 K T_{b}' - T_{b} = 0.1444 K T_{b}' = 0.1444 + 373.3 = 373.4444 K$

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18 g

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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}

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

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

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m o l^{- 1})

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1 k g

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, boiling points of pure water =

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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).

ΔTb=Kb × mwhere ΔTb = elevation in boiling pointKb = constantm = molalitymolality = no. of moles in 1 kg of solventm= mass givenMolar massMass of solvent(kg)=501801=0.277 mΔTb =0.52 × 0.2777 =0.1444 KTb'-Tb =0.1444 KTb' =0.1444 +373.3=373.4444 K