Question

Problem 1. i. There is one solution composed of solvent A and solute B. The concentration of B is given by CB, the molality is given by bb, and the density of the solution is p. If the molar mass of A and B is represented by MA and MB and the molar fraction of B in the solution is given by XB, i. derive the relation between xb and cb and xb and bb, respectively. ii. D-fructose (C6H1206) is dissolved in water to form a solution with mass fraction wb = 0.095, and density p = 1.0365 kg/L. By utilizing the equation you have derived above, try to calculate the molar fraction, concentration and molality.

Answer 1

ii)

Let total number of moles be 'n', moles of solvent A be n_A and that of solute B be n_B.

Let total weight be 'W', weight of solvent A be W_A and that of solute B be W_B.

Other variables are as defined in the question.

We know that the Weight is the product of molar mass and no. of moles.

Then,

$W_B=n_BM_B\qquad and \qquad W_B=n_BM_B\qquad(1)$

Total weight, W,

$W=n_BM_B+n_BM_B\qquad(2)$

Total Volume, V, will be Total weight divided by density i.e.,

$V=\frac{W}{\rho}=\frac{n_aM_a+n_bM_b}{\rho}\qquad(3)$

Further, let x_A be mole fraction of the Solvent A. Then, we know that the sum of mole fractions is 1. Thus, we will have,

$x_A=1-x_B\qquad(4)$

Further, we know that by definition of mole fraction,

$n_A=x_An\qquad and \qquad n_B=x_Bn\qquad(5)$

Finally we use the definition for molarity c_B as moles of solute divided by Volume

$c_B=\frac{n_B}{V}\\ \Rightarrow c_B=\frac{n_B}{\frac{n_AM_A+n_BM_B}{\rho}}\qquad\textrm{[Using (3)]}\\ \Rightarrow c_B=\frac{\rho x_Bn}{x_AnM_A+x_BnM_B}\qquad\textrm{[Using (5)]}\\ \Rightarrow c_B=\frac{\rho x_B}{(1-x_B)M_A+x_BM_B}\qquad\textrm{[Using (4)]}\\$

Thus, we have obtained c_B in terms of x_B.

Further, we know that Molality, b_b is given as Moles of solute per Kg of solvent. Thus,

$b_B=\frac{n_B}{W_B}\\ \Rightarrow b_B =\frac{n_B}{M_An_A}\qquad\textrm{[Using (1)]}\\ \Rightarrow b_B =\frac{x_Bn}{M_Ax_An}\qquad\textrm{[Using (5)]}\\ \Rightarrow b_B =\frac{x_B}{M_A(1-x_B)}\qquad\textrm{[Using (4)]}$

Thus, we have obtained b_B in terms of x_B

To put it together,

$c_B=\frac{\rho x_B}{(1-x_B)M_A+x_BM_B}\qquad\textrm{and}\qquad b_B =\frac{x_B}{M_A(1-x_B)}$

(iii)

Here, Molar Mass of D-Fructose C₆H₁₂O₆, M_B= 180 g/mol =0.180 kg/mol

Molar Mass of Water, H₂O, M_A= 18.0 g/mol =0.0180 kg/mol

In this question weight fraction, w_B is given.

$w_B=\frac{W_B}{W}\\ \Rightarrow w_B= \frac{n_BM_B}{\rho V}\qquad[\textrm{Using (1) and }W=\rho V] \\ \Rightarrow w_B=\frac{M_Bc_B}{\rho} \qquad[\textrm{Using definition of molarity}]\\ \Rightarrow c_B=\frac{w_B\rho}{M_B}\\ \Rightarrow c_B=\frac{(0.095)\times(1.0365 kg/L)}{0.180 kg/mol}\qquad\textrm{[Putting in the values]}\\ \Rightarrow c_B=0.547\;mol/L$

Now, we will use this value in the relation derived for c_B in terms of x_B

We have,

$c_B=\frac{\rho x_B}{(1-x_B)M_A+x_BM_B}\\ \Rightarrow 0.547=\frac{1.0365\times x_B}{(1-x_B)\times0.018+x_B\times0.180}\\ \Rightarrow 0.547=\frac{1.0365\times x_B}{0.018+x_B\times0.162}\\ \Rightarrow x_B=0.0104$

Thus, we have calculated the mole fraction. Now, we plug in this value in the relation for b_B. Thus, we will obtain,

$b_B =\frac{x_B}{M_A(1-x_B)}\\ \Rightarrow b_B=\frac{0.0104}{0.018\times(1-0.0104)}\\ \Rightarrow b_B=0.583\;mol/kg$

Thus, to put it together,

$x_B=0.0104;\quad c_B=0.547\;mol/L;\quad b_B=0.583\;mol/kg$