Question

Determine the freezing point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a melting point of 5.50°C and a freezing point depression constant of 4.90°C/m.

Answer 1

Step 1, - Calculate molality (m) of the naphthalene solution in benzene -

Given that,

( 78.8/ 128.16 ) = 0.6148 moles of naphthalene are dissolved in 722 ml ( or 722 x 0.877 gms = 633.194 gms ) of benzene.

Since molality (m) of a solution is = Number of moles of solute / Mass of solvent in kg

molality (m) of the given solution of naphthalene in benzene = (0.6184 x 1000) / 633.194

........................................................ ..............................................= 0.9766 m

Step 2, - Apply the relation,

T_f  = K_f .m.............[where T_f represents depression in freezing point of the solvent ( ie. Freezing point of pure solvent - Freezing point of solution ) , K_f is freezing point depression constant , given = 4.90^o C / mol , & m = molality of solution = 0.9766 m ,calculated as above.

Substituting the values in the relation we get ,

T_f  = 4.90 x 0.9766

........ = 4.78 ^o C

Step 3, -

Now T_f  = (Freezing point of pure benzene) - (Freezing point of solution)

........4.78 = 5.5 - freezing point of solution

Therefore , Freezing point of solution = ( 5.5 - 4.78 )^o C

............................................................. = 0.72^o C