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The freezing point ("in"^(@)"C") of a so...

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

A

`-2.3xx10^(-2)`

B

`-5.7xx10^(-2)`

C

`-5.7xx10^(-3)`

D

`4.0xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`"For" K_(3)[Fe(CN)_(6)]i=4`
`DeltaT_(f)=ixx(K_(f)W_("solute")xx1000)/(m_("solute")xxw_("solvent"))`
`DeltaT_(f)=4xx1.86xx(0.1xx1000)/(329xx100)`
`thereforeT_(f)("solution")=0.0023=-2.3xx10^(-2@)"C"`
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Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solution molecules ae added to get homogeneous solution. These are called colligative properties. Application colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given: Freezing point depression of water (K_(f)^("water"))=1.86K kg "mol"^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol")=2.0K kg "mol"^(-1) Boiling point elevation constant of water (K_(b)^("water")=0.52K kg "mol"^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol")=1.2K kg "mol"^(-1) Standard freezing point of water =273K Standard freezing point of ethanol =155.7K Standard boiling point of water =373K Stadndard boiling point of ethanol =351.5K Vapour pressure of pure water =32.8mm Hg Vapour pressure of pure ethanol =40mm Hg Molecular weight of water =18 g "mol"^(-1) Molecular weight of ethanol =46g "mol"^(-1) In aswering the following questions, consider the solution to be ideal dilute solutions and solutes to be non volatile and non-dissociative. The freezing point (in .^(@)C ) of a solution containing 0.1g of K_(3)[Fe(CN)_(6)] (Mol. Wt. 329 ) in 100g of water (K_(f)=1.86K kg "mol"^(-1) is :

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

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 :

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

The freezing point (.^(@)C) of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] (molecular weight 329) on 100 g of water (K_(f)=1.86 K kg mol^(-1))

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) )

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