Home
Class 12
CHEMISTRY
45 g fo ethylene glycol (C(2)H(6)O(2)) i...

45 g fo ethylene glycol `(C_(2)H_(6)O_(2))` is mixed with 600 g of water. Calculate (a) the freezing point depression and (b). The freezing point of the solution.

Text Solution

Verified by Experts

Depression in freezing point is realed to the molality, therefore the molality of the solution with respect to ethylene glycol,
`DeltaT_(f)=K_(f)m`
Mole of ethylene glycol = `45 gxx(1 mol)/(62g) =0.73 mol`
Mass of water in kg =`600 xx (1 kg)/(1000 g)=0.60 kg`
Hence, molality of ethylene glycol
`=(0.73 mol)/(0.60 kg)=1.20 mol kg`
Therefore,freezing point depression
`(DeltaT_(f)=1.86 K kg mol^(-1))xx(1.2 mol kg^(-1) )=2.2 K`
Freezing point of the aqueous solution
`T_(f) =T_(f)^(@) - Delta_(f)T`
`273.15 K -2.2 K =270.95 K`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Solved Examples|40 Videos

  • SOLUTIONS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Linked Comprehension)|58 Videos

  • SOLID STATE

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.2 (Objective)|9 Videos

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|2 Videos

Similar Questions

Explore conceptually related problems

45 g of ethylene glycol C_(2)H_(6)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

45 g of ethylene glycol C_(2)H_(4)O_(2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given K_(f)=1.86 K kg mol^(-1) .

The freezing point of the H_2O is :

45 g of ethylene glycol (C_(2) H_(6)O_(2)) is mixed with 600 g of water. The freezing point of the solution is (K_(f) for water is 1.86 K kg mol^(-1) )

An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_(2)H_(4)(OH)_(2)] and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072g mL^(-1) then what shall be the molarity of the solution?

In which of the following solution the depression in freezing point is lowest ?

The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol (CH_(2)OH-CH_(2)OH) is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol (CH_(3)OH) , a fairly volatile liquid that boils only at 65^(@)C , is sometimes used as an antifreeze in automobile radiators. 620 g glycol is added to 4 kg water in the radiator of car. What amount of ice will separate out at -6^(@)C ?

The boiling point elevation and freezing point depression of solutions have a number of partical applications. Ethylene glycol (CH_(2)OH-CH_(2)OH) is used in automobile radiatiors as an antifreeze because it lowers the freezing point of the coolant. The same substance also helps to prevent the radiator coolant from boiling away by elevating the boiling point. Ethylene glycol has low vapour pressure. We can also use glycerol as an antifreeze. In order for the boiling point elevation to occur, the solute must be non-volatile, but no such restriction applies to freezing point depression. For example, methanol (CH_(3)OH) , a fairly volatile liquid that boils only at 65^(@)C , is sometimes used as an antifreeze in automobile radiators. Which of the following is a better reagent for depression in freezing point but not for elevation in boiling point?

When 20 g of napthanoic acid (C_(11)H_(8)O_(2)) is dissolved in 50 g of benzene (K_(f)=1.72 K kg/mol) a freezing point depression of 2 K is observed. The van't Hoff factor (i) is