Home
Class 12
CHEMISTRY
Calcualate the amount of NaCl which must...

Calcualate the amount of `NaCl` which must be added to `100 g` water so that the freezing point, depressed by `2 K`. For water `K_(f)` =`1.86 K kg mol^(-1)`.

Text Solution

Verified by Experts

`NaCl` is a strong electrolyte. It is completely dissociated in solution.
Degree of dissociation, `alpha=1`
`NaClhArrunderset((n=2))(Na^o++Cl^(ө))`
Number of particles after dissociation =`1+(n-1)alpha`
`=1+(2-1)xx1`
`=2`
`DetlaT_(obs)/DetlaT_(theo)="Number of particles after dissociation"/"Number of particles when there is no dissociation"`
`2/DetlaT_(theo)=2`
ot `DetlaT_(theo)=1`
Let W g of NaCl be dissolved in `100 g` of water.
So `DetlaT_(theo)=(1000xxK_(f)xxW_(2))/(W_(1)xxMw_(1))`
or`W_(2)=(DetlaT_(theo)xxW_(1)xxMw_(1))/(1000xxK_(f))`
`(1xx100xx58.5)/(1000xx1.86) = 3.145 g`
Promotional Banner

Similar Questions

Explore conceptually related problems

Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_(f) " for water = " 1.86 " K kg "mol^(-1))

In a cold climate water gets frozen causing damage to radiator of a car. Ethylene glycol is used as an anifreezing agent. Calculate the amount of ethylene glycol to be added to 4 kg of water to prevent it from freezing at -6^(@) . ( K_(f) for water = 1.85 kg mol^(-1) )

How many ethyl alcohol must be added to 1.00 L of water so that the solution will not freeze at -4^(@)F ?

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at - 6^(@)C will be ( K for water = 1.86 "K kg mol"^(-1) , and molar mass of ethylene glycol = 62 g "mol"^(-1) )

What mass of ethylene glycol (molar mass = 62.0 g mol^(-1) ) must be added to 5.50 kg of water to lower the freezing point of water from 0^(@)C to - 10.0^(@) C ? (K_(f) "for water " = 1.86 k " kg mol"^(-1))

What do you understand by the term that K_(f) for water is 1.86 K kg mol^(-1) ?

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 constant of water (K_(f)^(water)=1.86 K kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=2.0 K kg mol^(-1)) Boiling point elevation constant of water (K_(b)^(water))=0.52 K kg mol^(-1)) Boiling point elevation constant of ethanol (K_(b)^(ethanol))=1.2 K kg mol^(-1)) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9 . The boiling point of this solution is

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 constant of water (K_(f)^(water)=1.86 K kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=2.0 K kg mol^(-1)) Boiling point elevation constant of water (K_(b)^(water))=0.52 K kg mol^(-1)) Boiling point elevation constant of ethanol (K_(b)^(ethanol))=1.2 K kg mol^(-1)) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution M is