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

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 :

A

`-2.3 xx10^(-2)`

B

`-5.7 xx 10^(-2)`

C

`-5.7 xx 10^(-3)`

D

`-1.2 xx 10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Total mass of solution `=(1000+120)=1120 g`
Total volume of solution `= ("Mass")/("Density")=(1120)/(1.15)`
`= 973.91 mL`
Molarity `M_(f)=(W)/(m)xx(1000)/(V(mL))`
`=(120)/(60)xx(1000)/(973.91)=2.05 M`
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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

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 100g 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))

What will be the freezing point ("in"^(@)C) of solution obtained by dissolving 0.1 g potassium ferricyanide (mol wt = 329) in 100 g water. If K_(f) for water is 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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