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Calculate the freezing point of a soluti...

Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g `mol^(-1)`) in 250 g of water . (`K_(f)` of water = `1.86 K kg mol^(-1)`)

A

271.67 K

B

270.67 K

C

274 K

D

270 K

Text Solution

Verified by Experts

The correct Answer is:
B
Freezing point of solvent = Molal
depression constant ` xx ("Mass solute")/("Molar mass ")xx(1000)/("Mass of solvent")`
`DeltaT_(f) = K_(f) xx (W_(B))/(M_(B)) xx (1000)/(W_(A))`
` = 1.86xx (60)/(180) xx (1000)/(250)`
=2.48K
Freezing point = 273.15 K -2.481
=270.67K.
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Knowledge Check

  • The boiling point of solution containing 68.4 g of sucrose ("molar mass" = 342 g mol^(-1)) in 100 g of water is (K_(b) "for water" = 0.512 K. kg mol^(-1))

    A
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    C
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    D
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    A
    34.2g
    B
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    C
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    D
    72g

  • The freezing point (in .^(@)C) of a solution containing 0.1g of K_(3)[Fe(CN)_(6)] (Mol. wt. 329 ) in 100 g of water (K_(f) = 1.86 K kg mol^(-1)) is :

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    B
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    C
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    D
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