Answer to Question #139489 in General Chemistry for Princess Mariano

Question

2. Calculate the molar mass of the solute in a solution of 2.47 g of a mo electrolyte in 100 g of acetic acid (CH3COO). The solution freezes at 1.3 °C below the normal freezing point of pure acetic acid. (Kf for CH3COOH=3.9°C/m)

Answer 1

2.47 g of a mo electrolyte

100 g of acetic acid (CH3COO).

The solution freezes at 1.3 °C

(Kf for CH3COOH=3.9°C/m)

Therefore :

2.47/ 100 =0.0247moles

1 mole. = 3.9 ^O_C

? = 1.3 ^O_C

= 1.3/3.9 = 0.3moles

0.0247moles = 2.47g

0.3 moles = ((0.3)(2.47))/0.0247

= 30g