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

Answer:

62.06 g/mol

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=$3.33^{\circ}$

$\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}$

$M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}$

$M_B=62.06 g/mol$

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

Answer 2

Answer:

Molar mass of the unknown liquid = 62.07g/mol

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