0.10 mol of molecule A is added to 1.0 mol pf molecule B.The initial volume of B was 1.1L and the final volume of the solution is 1.039L. Assuming that the partial molar volume of B is exactly 50% greater than that of A, find the partial molar volumes of both A and B.
To find the partial molar volumes of A and B, we can use the concept of the molar volume, which is the volume occupied by one mole of a substance.
First, let's calculate the amount of substance B that was present initially. We are given that the initial volume of B is 1.1L, and the final volume of the solution is 1.039L. Therefore, the change in volume of B is:
ΔV(B) = Final volume - Initial volume
= 1.039L - 1.1L
= -0.061L
Since the change in volume is negative, it means that the volume of B has decreased. Now, we can calculate the number of moles of substance B initially:
n(B) = (Change in volume of B) / (Molar volume of B)
The molar volume of B is given as 50% greater than that of A, so we can express it as:
Molar volume of B = (1 + 0.50) * (Molar volume of A)
= 1.50 * (Molar volume of A)
Substituting this into the equation to calculate n(B), we have:
n(B) = (Change in volume of B) / (1.50 * Molar volume of A)
Plugging in the values we have so far, we get:
n(B) = -0.061L / (1.50 * Molar volume of A)
Next, we know that 0.10 mol of A is added to the solution. Therefore, the total moles of B in the solution is:
n(B)total = n(B)initial + n(B)added
= n(B) + 0.10 mol
We can rearrange this equation to solve for n(B)initial:
n(B)initial = n(B)total - 0.10 mol
Now, substitute the expression for n(B) into the equation:
(n(B)total - 0.10 mol) = -0.061L / (1.50 * Molar volume of A)
Rearranging the equation to solve for the molar volume of A, we get:
Molar volume of A = -0.061L / (1.50 * (n(B)total - 0.10 mol))
Finally, to find the molar volume of B, we can substitute the molar volume of A into the equation:
Molar volume of B = 1.50 * Molar volume of A
Now, we can calculate the partial molar volumes of A and B using the equation we derived.