Question

For a binary solution, find expressions for the mole fractions x_B and x_A as functions of the solute molality m_B.

Answer 1

total mol = mol of B + mol of A

xb = mol of B / total mol

xb = mol of B / (mol of A + mol of B)

xa = mol of A / (mol of A + mol of B)

then...

if we assume Mb = mol of B / kg of solvent

and assume solvent = A

then

Mb = mol of B / kg solvent

kg of solvent = kg of A = mass of A = mol of A * MW of A

so

Mb = mol of B / (mol of A * MW of A)

solve for mol of B

mol of B = Mb* ((mol of A * MW of A)) =

substitute

xb = mol of B / (mol of A + mol of B)

xb =Mb* ((mol of A * MW of A)) / (mol of A + Mb* ((mol of A * MW of A))

for A:

xa = mol of A / (mol of A + Mb* ((mol of A * MW of A))