Q. 65.0( 1 Vote )

# Calculate the freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be 90 % ionized. [Given: Molar mass Br = 80 g/mol, Kf water = 1.86 K kg / mol]

**OR**

Calculate the molality of ethanol solution in which the mole fraction of water is 0.88.

Answer :

Given:

Mass of HBr given = 8. 1g

Mass of water = 100g

Percentage of ionization of acid = 90% so, α = 0. 90

Molar mass of HBr = 81 g

K_{f} = 1. 86 k kg/mol

As we know:

HBr dissociates to:

HBrH^{+} + Br^{-}

To find the Van’t Hoff factor (i):

i = 1 +

__The formula__ __to find the__

= iK_{f}m

Substitute i, K_{f},m in the equation

i = (1+α)

where,

α = 0.9

= (1 +

= (1 +

= 3. 53

T^{0}_{f}- = 0

= T^{0}_{f}-T^{1}_{f}

T_{f} = -3.534

__Conclusion:__

Freezing point of the given solution is -3.534 °C

**OR**

Given that,

The mole fraction of water = 0. 88

Therefore, the mole fraction of ethanol will be = 1 - 0. 88 = 0. 12

The formula to find the mole fraction of a substance is =

0. 12 = (for ethanol) ……… (1)

0. 88 = (for water)………… (2)

Divide equation two by equation 1

7. 333 =

No. of moles of water in 1000g of water = = 55. 55

Therefore no. of moles of ethanol = = 7. 575

Molality =

__Conclusion:__

So, molality = 7. 575 mol/kg.

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