Question #d6d89
1 Answer
Explanation:
So, lanthanum(III) bicarbonate,
#"La"("HCO"_3)_text(3(s]) -> "La"_text((aq])^(3+) + 3"HCO"_text(3(aq])^(-)#
Now, the mean molality of the solution can be thought of as the average molality of the two ions formed when the compound dissociates
#m_(pm) = [m_(+)^(nu_+) + m_(-)^(nu_(-))]^(1/nu)" "# , where
Now, because no information was given about the mass of solvent, I will assume that the concentration you provided is actually 0.002 molal, not molar.
So, every formula unit of lanthanum(III) bicarbonate produces one lanthanum(III) cation and three bicarbonate anions, so you know that
#nu_(+) = 1" "# #" "nu_(-) = 3" "# #" "nu = v_(+) + nu_(-) = 4#
This means that the mean molality of the solution will be
#m_(pm) = [(0.002)^1 + (3 xx 0.002)""^3]^(1/4)#
#m_(pm) = (0.002 + 0.006""^3)""^(1/4) = color(green)("0.211 molal")#
The ionic strenght of the solution is defined as
#I = 1/2 * sum_i(m_i * z_i^2)" "# , where
In your case, the ionic strenght of the solution will be
#I = 1/2 * [0.002 * ("3+")^(2) + 0.006 * ("1-")^2]#
#I = 1/2 * (0.018 + 0.006) = color(green)(0.012)#