# 8g of NaOH is dissolved in 18g of ${{H}_{2}}O$. Mole fraction of NaOH in solution and molality (in mol/kg) of the solutions respectively are:

(a) 0.167, 11.11

(b) 0.2, 22.20

(c) 0.2, 11.11

(d) 0.167, 22.20

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**Hint:**In this first we have to find the no of moles of NaOH and ${{H}_{2}}O$ and then we can find the mole fraction of NaOH by using the formula as Mole fraction of NaOH= $\dfrac{{{n}_{NaOH}}}{{{n}_{NaOH}}+{{n}_{{{H}_{2}}O}}}$ and we know the mass of the water and can easily find the molality of the solution by using the formula as; Molality =$\dfrac{\text{no of moles of the solute}}{\text{ total mass of the solvent in kg}}$. Now solve it.

**Complete step by step answer:**

First of all, let’s discuss mole fraction and molality. By the term mole fraction we mean the ratio of the number of a particular component to the total number of moles of the solution (i.e. solute or solvent). If the substance A is dissolved in solvent B and ${{n}_{A}}$ and ${{n}_{B}}$ are the mole fractions of the solute A and solvent B , then;

Mole fraction of solute A=$\dfrac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}$

Mole fraction of solvent B= $\dfrac{{{n}_{B}}}{{{n}_{A}}+{{n}_{B}}}$

And by the term molality we mean the no of moles of the solutes to the total mass of the solvent in kilograms. i.e.

Molality =$\dfrac{\text{no of moles of the solute}}{\text{ total mass of the solvent in kg}}$ -----------(A)

Now, considering the numerical;

We can calculate the moles of NaOH by using the formula as;

Moles of NaOH =$\dfrac{given\text{ }mass}{molecular\text{ }mass}$ ---------(1)

Given mass of NaOH= 8g

Molecular mass of NaOH= 23+16+1= 40

Put these values in equation(1), we get;

Moles of NaOH =$\dfrac{8}{40}$

=0.2

Similarly,

Given mass of ${{H}_{2}}O$= 18g

Molecular mass of ${{H}_{2}}O$= 2+16= 18

Put these values in equation(1), we get;

Moles of NaOH =$\dfrac{18}{18}$

= 1

Mole fraction of NaOH= $\dfrac{{{n}_{NaOH}}}{{{n}_{NaOH}}+{{n}_{{{H}_{2}}O}}}$

Put the values of number of moles of NaOH and ${{H}_{2}}O$ in it, we get;

Mole fraction of NaOH= $\dfrac{0.2}{0.2+1}$

=$\dfrac{0.2}{1.2}$

= 0.167

Now, calculating the molality of the solution using the equation (A) as;

Molality =$\dfrac{\text{no of moles of the NaOH}}{\text{ total mass of the }{{\text{H}}_{2}}\text{O in kg}}$---------(2)

No of moles of NaOH= 0.2

Mass of water=18g

= $\dfrac{18}{1000}$ kg ( 1kg=1000g)

Put these values in equation (2) we get:

Molality = $\dfrac{0.2\times 1000}{18}$

= 11.11m

**Hence, option(a) is correct.**

**Note:**The sum of the mole fraction of the solute and solvent is always equal to one and can never be greater than one but can be less than one and molality of the solution is independent of the temperature and depends only on the mass of the solvent.