Question

can y’all help me please?

The freezing point of water, H2O, is 0.000 °C at 1 atmosphere. K(water)--1.86 °C/m In a laboratory experiment, students synthesized a new compound and found that when 12.51 grams of the compound were dissolved in 213.6 grams of water, the solution began to freeze at -1.813 °C. The compound was also found to be nonvolatile and a non-electrolyte What is the molecular weight they deterfnined for this compound ? g/mol

Answer 1

59.50 g/mol

Molality (m) = mas of new compound (g)/(M_wt of compound) (mass of water in kg) rightarrow Therefore m = 12.51 g/(M_wt)(213.6 times 10^- 3 kg) rightarrow for non - volatile and non - electrolyte compound l = Van't Hoff factor = 1 rightarrow Delta t_f = freezing point of h_2 o - freezing point if compound solution = 0.000 degree c - (- 1.813 degree c) Delta t_f = 1.813 degree c rightarrow now, from the equation Delta t_f = k_f. m. i 1.831 degree c = (1.86 degree c/mol/kg) (12.51 2/m_wt times 213.6 times 10^- 3 kg) (1) Therefore m_wt = (1.86) (12.51) (1)/(1.831) (213.6 times 10^- 3) g/mol = 59.50 g/mol