Question: How many grams of sodium chloride per kilogram of water are present in a 2.7 m aqueous solution?
The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all the components of the solution. \(\begin{array}{l}Mole fraction =\frac{Number of moles of solute}{Number of moles of solvent}\end{array} \) \(\begin{array}{l}mole fraction = \frac{n_{B}}{n_{A}+n_{B}}\end{array} \) nB= number of moles of solute nA = Number of moles of solvent Molality is defined as the total moles of a solute contained in a kilogram of a solvent. \(\begin{array}{l}Molality = \frac{Moles of solute}{Moles of solven}\end{array} \) \(\begin{array}{l}Molality = \frac{n_{B}}{W_{A}(in kg)}\end{array} \) The relation between the mole fraction and molality is given as: Consider a binary solution component with A as solvent and B as solute. Let, Mole fraction of solvent = xA Mole fraction of solute = xB Number of moles of solvent = nA Number of moles of solute = nB Mass of solvent = WA Mass of solute = WB Molar mass of solvent = MA The molar mass of solute = MB Mole fraction of solute = xB = \(\begin{array}{l}\frac{n_{B}}{n_{B}+n_{A}}\end{array} \) Mole fraction of solvent = xA = \(\begin{array}{l}\frac{n_{A}}{n_{B}+n_{A}}\end{array} \) (eq.2)Dividing eq 1 and 2, \(\begin{array}{l} \frac{X_{B}}{X_{A}} = \frac{\frac{n_{B}}{n_{B}+n_{A}}}{\frac{n_{A}}{n_{B}+n_{A}}}\end{array} \) Therefore, \(\begin{array}{l}frac{x_{B}}{x_{A}}=frac{frac{W_{B}}{M_{B}}}{frac{W_{A}}{M_{A}}}=frac{W_{B}}{W_{A}}times frac{M_{A}}{M_{B}}\end{array} \) Therefore, \(\begin{array}{l}frac{x_{B}times 1000}{x_{A}times M_{A}}=frac{W_B}{M_B}times frac{1000}{W_A}=frac{n_Btimes 1000}{W_A}\end{array} \) = Molality = mTherefore, Molality = m = \(\begin{array}{l}frac{x_{B}times 1000}{x_{A}times M_{A}}=frac{x_Btimes 1000}{(1-x_B)times M_A}\end{array} \) Solved Examples1. Calculate the mole fraction of ammonia in a 3.00 m solution of ammonia in water. Solution: Let us assume 1 kg of solvent that is 1000 g of water The molecular mass of water = 18.015 g mol–1 Moles of water = 1000/18.015 Moles of water = 55.402 For 1 kg of H2O, there are 3 moles of NH3 \(\begin{array}{l}Mole fraction = frac{number of moles of solute}{Number of moles of solvent}\end{array} \) \(\begin{array}{l}Mole fraction = frac{3}{3 + 55.402}\end{array} \) Mole freaction = 0.0514 2. Calculate the molality of the solution when 70.128 grams of NaCl salt is dissolved in 2 grams of water. Solution: Number of moles of NaCl = 70.128 g/ 58.44 Molar mass of NaCl = 58.44 g/mol Number of moles of NaCl = 1.2 moles Molality = 1.2/2 Molality = 0.6 mol/kg Check out the video given below to know more about solvent and solute  Further Reading
Answer Hint: Mole fraction of any solution is defined as the number of moles in one component upon the total number of moles in all components. The components may be the solute or the solvent. The molality of any solution is the number of moles of solute dissolved per kilogram of the solvent used in a solution. Complete answer: We have been given to find the relationship between the mole fraction of a solute$\left( {{X}_{A}} \right)$and its molality (m) given that the molar mass of solvent is 100 (g/mol). We have the expression of mole fraction as number of moles in component A upon the total number of moles in all components A solute and solvent as ${{X}_{A}}=\dfrac{{{n}_{a}}}{{{n}_{a}}+{{n}_{solvent}}}$ , where n is the number of moles. From the expression of molality we have, $m=\dfrac{{{n}_{A}}\times 1000}{{{n}_{solvent}}\times 100}$ , so, $m=\dfrac{10\times {{n}_{a}}}{{{n}_{solvent}}}$ .Now, dividing the mole fraction expression with molality we have,$\dfrac{{{X}_{A}}}{m}=\dfrac{\dfrac{{{n}_{a}}}{{{n}_{a}}+{{n}_{solvent}}}}{\dfrac{10\times {{n}_{a}}}{{{n}_{solvent}}}}$ $\dfrac{{{X}_{A}}}{m}=\dfrac{{{n}_{a}}\times {{n}_{solvent}}}{10{{n}_{a}}\times ({{n}_{a}}+{{n}_{solvent}})}$Rearranging the above equation and solving for molality we have,$m=\dfrac{10{{X}_{A}}({{n}_{a}}+{{n}_{solvent}})}{{{n}_{solvent}}}$ , as mole fraction of solvent is$\dfrac{{{n}_{solvent}}}{({{n}_{a}}+{{n}_{solvent}})}={{X}_{solvent}}$ , therefore, $m=\dfrac{10{{X}_{A}}}{{{X}_{solvent}}}$ As mole fraction of other component (solvent) can be obtained as $1-{{X}_{A}}$, therefore,$m=\dfrac{10{{X}_{A}}}{1-{{X}_{A}}}$Hence, the relationship between mole fraction of a solute$\left( {{X}_{A}} \right)$and its molality (m) is$\dfrac{10({{X}_{A}})}{(1-{{X}_{A}})}$.So, option D is correct. Note: The mole fraction of all the components is equal to unity therefore the expression, ${{X}_{solvent}}=1-{{X}_{A}}$. Mole fraction has no unit as it is the fraction. As molality is expressed in kilograms, therefore in the formula of molality, the volume is divided by the factor of 1000, to make it into kilograms. |
What is the relationship between mole fraction of a solute and its molality if molar mass of solvent is 100?
Pos Terkait
Copyright © 2024 ketajaman Inc.
