Question

Dissolve a substance of 0.900 g in 60.0 g water, to achieve the freezing point drop of 0.150 K, the molecular weight of the material is (water Kf = 1.86 K•kg/mol) :​

Answer 1

THE MOLECULAR WEIGHT OF UNKNOWN SUBSTANCE DISSOLVED IN 60 gm OF WATER IS = 186 gm/mol

IT IS GIVEN THAT 0.900 gm OF UNKNOWN SUBSTANCE IS DISSOLVED IN 60 gm OF WATER

• The depression in freezing point is a colligative property that occurs when the molecules of solute are added into the solvent results in the lowering of freezing point and is given by ,

                                        ΔTf= Kf × m

• where ΔTf= Tf(solvent)-Tf(solution) and m is molality

• According to given question,

• Weight of unknown substance dissolved= 0.90gm

• Weight of solvent(water)= 60.0 gm

• Depression in freezing point , ΔTf= 0.150 K              Kf=1.86 K.KG/MOL

• LET M be the molecular weight of unknown substance

• molality , m= $\frac{0.900}{M}$ ×$\frac{1000}{60.0}$ =$\frac{15}{M}$

• ΔTf= Kf × m = 1.86 × $\frac{15}{M}$ = 0.150

HENCE, the molecular weight of unknown substance is

                                         186 g/mol

THANKS!!

                                     

Answer 2

Thus the molecular weight of unknown substance is 186 g/mol

Explanation:

The lowering of freezing point and is given by "ΔTf" = Kf × m

Where

• ΔTf = Tf (solvent) - Tf (solution)
• m = molality

According to given question

Weight of unknown substance dissolved= 0.90 g

Weight of solvent(water) = 60.0 g

Depression in freezing point "ΔTf" = 0.150 K              

Kf = 1.86 K.Kg / mol

LET M be the molecular weight of unknown substance

Molality , m = 0.9 / M × 1000 / 60 = 5 / M

ΔTf =  Kf × m = 1.86 × 15 / M = 0.150

Thus the molecular weight of unknown substance is 186 g/mol