Question

25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is

0.57 °C what is the molecular weight of the compound assume no molecular disassociation upon
dissolution Kf equals 1.36 °C/m
O 119 g/mol
90 g/mol
0 60 g/mol
238 g/mol

Answer 1

a) 119 g/mol

Step-by-step explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

`\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=(\bigtriangleup T_f)/(K_f)\\\\=(0.57\textdegree C)/(1.36\textdegree C)\\\\=0.4191\ mol/Kg\\\\`

#We use the molality above to calculate the molar mass:

`m=(0.4191\ mol)/(1\ Kg)=(25\ g)/(0.5\ Kg)\\\\\therefore 1 \ mol=(25\ g)/(0.5\ Kg)*(1\ Kg)/( 0.4191)\\\\=119.3033\approx 119\ g/mol`

Hence, the molar mass of the compound is 119 g/mol