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Question

Question: Mole fraction of solute in a 1.00 molal aqueous solution: (A) 1.7700 (B) 0.1770 (C) 0.0177 (...

Mole fraction of solute in a 1.00 molal aqueous solution:
(A) 1.7700
(B) 0.1770
(C) 0.0177
(D) 0.0844

Explanation

Solution

Molal is a term that comes from Molality. Molal means a solution that contains one mole of solute in 1000g or 1kg of solvent. Mole fraction is the number of moles of a certain specific component to the total number of moles present in the mixture. We can find the mole fraction of solute using the following formula
Mole fraction of solute xA=nAnA+nB{x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}

Complete step by step solution:
In order to find the mole fraction of a solute, we must first know what a mole fraction is. Mole fraction is defined as the number of moles of a certain component present in the mixture to the total number of moles of all components in the mixture.
Let us consider nA{n_A}and nB{n_B}as the number of moles of solute and solvent respectively. We can calculate the mole fraction of solute and solvent by the formula given below:
Mole fraction of solute, xA=nAnA+nB{x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}
Mole fraction of solvent, xB=nBnA+nB{x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}
Sum of the mole fraction of solute and solvent will be unity,
i.e. xA+xB=1{x_A} + {x_B} = 1
In the question, it has given about one molal aqueous solution which means a solution that contains one mole of solute in 1000g or 1kg of solvent.
Therefore, the number of moles of solute present is 1.
Now, we have to find the moles of solvent. Since which solvent is present is not given, we can consider the solvent as the water.
Molar mass of the solvent = 1000g
Number of moles of solvent i.e. water =1000Molarmassofwater{\rm{ = }}\dfrac{{{\rm{1000}}}}{{{\rm{Molar mass of water}}}}
Number of moles of solvent i.e. water =100018=55.55{\rm{ = }}\dfrac{{{\rm{1000}}}}{{18}} = 55.55
Therefore, the number of moles of solvent is 55.55
Mole fraction of solute is given by
xA=nAnA+nB{x_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}
Mole fraction of the solute is xA=11+55.55=156.55=0.01768{x_A} = \dfrac{1}{{1 + 55.55}} = \dfrac{1}{{56.55}} = 0.01768
The value 0.01768 is approximated to 0.0177
Therefore, xA=0.01768=0.0177{x_A} = 0.01768 = 0.0177

Hence the correct answer is (C) 0.0177

Additional information:
Let us now discuss what are Molarity, Normality and Molality
- Molarity is defined as the number of moles of solute dissolved in one litre (or 1000ml) of solution. It is represented by M.
- Normality is defined as the number of gram-equivalent of solute dissolved in one litre (or 1000ml) of solution. It is represented by N.
- Molality is defined as the number of moles of solute dissolved in one kilogram (or 1000g) of solvent. It is represented by m.

Note: - Mole fraction is a dimensionless quantity.
- In order to calculate the mole fraction of any compound, we have to convert the values given into moles.
- If a specific solvent is not given, we can take the solvent as water. We can directly use the value 55.55 as the number of moles of solvent.