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 =[tex]W_B[/tex]=10 g
Given mass of solvent=[tex]W_A[/tex]=90 g
[tex]k_f=1.86^{\circ}C/m[/tex]
Freezing point of solution =-3.33[tex]^{\circ}[/tex]C
Freezing point of solvent =[tex]0^{\circ}[/tex]C
Change in freezing point =Depression in freezing point
=Freezing point of solvent - freezing point of solution=0+3.33=[tex]3.33^{\circ}[/tex]
[tex]\Delta T_f=\frac{W_B\times K_f\times 1000}{W_A\times M_B}[/tex]
[tex]M_B=\frac{10\times 1.86\times 1000}{3.33\times 90}[/tex]
[tex]M_B=62.06 g/mol[/tex]
Hence, molar mass of unknown liquid is 62.06g/mol.
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
