When HClO2 is dissolved in water, it partially dissociates according to the equation

HClO2 → H+ + ClO2-. A solution is prepared that contains 5.850 g of HClO2 in 1.000 kg of water. Its freezing point is -0.3473 °C. Calculate the fraction of HClO2 that has dissociated.

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

Explanation:

Recall that , depression present in freezing point is calculated with the formulae = solute particles Molarity x KF

0.3473 = m * 1.86

Solving, m = 0.187 m

Moles of HClO2 = mass / molar mass = 5.85 / 68.5 = 0.0854 mol

Molality = moles / mass of water in kg = 0.0854 / 1 = 0.0854 m

Initial molality

Assuming that a % x of the solute dissociates, we have the ICE table:

                 HClO2         H+    +   ClO2-

initial concentration:       0.0854                    0             0

final concentration:      0.0854(1-x/100)   0.0854x/100   0.0854x / 100

We see that sum of molality of equilibrium mixture = freezing point molality

0.0854( 1 - x/100 + x/100 + x/100) = 0.187

2.1897 = 1 + x / 100

x = 100 * 1.1897 = 118.97 %, which is > 100 meaning that all of the HClO2 dissociates

answer; the numbers of valence electrons in the acetic acid molecule (ch3co2h) is 24;

c

explanation: