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

[tex]\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\[/tex]

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

[tex]m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol[/tex]

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

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

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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