Question

A solution containing 10.0 g of an unknown liquid and 90.0 g water has a freezing point of -3.33 °C. Given Kf = 1.86 °C/m for water, the molar mass of the unknown liquid is ________ g/mol.

Answer 1

Molar mass of the unknown liquid = 62.07g/mol

Step-by-step explanation:

Formula is given as

Molality of solution = freezing point of water ÷ Kf

Where

Freezing point of water = -3.33 °C

Kf = in the question, water is the solvent = 1.86 °C/m

Molality of the Solution = 3.33 °C ÷ 1.86 °C/m

= 1.79mol/kg

Next step is to find the number of moles of the solute

Number of moles of solute = Molality of solution × kg of water

Water = 90g

1000g = 1kg

90g of water = 90÷1000

= 0.09kg of water

Number of moles of solute = 1.79mol/kg × 0.09kg of water

= 0.1611mole of solute

Molar mass of the unknown liquid =

Mass of the unknown liquid ÷ Number of moles of solute

Mass of the unknown liquid = 10g

Number of moles of solute = 0.1611mole

Molar mass of the unknown liquid = 10g ÷ 0.1611mole

= 62.07g/mole

Answer 2

62.06 g/mol

Step-by-step explanation:

We are given that a solution containing 10 g of an unknown liquid and 90 g

Given mass of solute =
`W_B`=10 g

Given mass of solvent=
`W_A`=90 g

`k_f=1.86^(\circ)C/m`

Freezing point of solution =-3.33
`^(\circ)`C

Freezing point of solvent =
`0^(\circ)`C

Change in freezing point =Depression in freezing point

=Freezing point of solvent - freezing point of solution=0+3.33=
`3.33^(\circ)`

`\Delta T_f=(W_B* K_f* 1000)/(W_A* M_B)`

`M_B=(10* 1.86* 1000)/(3.33* 90)`

`M_B=62.06 g/mol`

Hence, molar mass of unknown liquid is 62.06g/mol.