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.

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 =

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_fm

Substitute i, K_f,m in the equation

= (1 +

= (1 +

= 3. 53

T⁰_f- = 0

= T⁰_f-T¹_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.