Question

Calculate (a) molality (b) molarity and (c) mole fraction of $KI$ if the density of $20\%$ (mass/mass) aqueous $KI$ is $1.202{{ }}g/mL$.

Answer 1

The concentration of a compound can be expressed in different ways in chemistry. Concentration is defined as the amount of solute present in a unit volume of solution. Different terms used for expressing the concentration are Molarity, Molality, Normality, Mole fraction, Parts per million(ppm) etc.

Formula Used:
The Molarity of a solution The concentration of a compound can be expressed in different ways in chemistry. Concentration is defined as the amount of solute present in a unit volume of solution. Different terms used for expressing the concentration are Molarity, Molality, Normality, Mole fraction, Parts per million(ppm) etc.
The Molarity of a solution is represented by the formula:
$M = \dfrac{{No.\,of\,moles\,\,of\,solute}}{{Volume\,of\,solution(in\,L)}}$
The Molality of a solution is represented by the formula:
$m = \dfrac{{No.\,of\,moles\,\,of\,solute}}{{mass\,of\,solvent(in\,Kg)}}$
The density of a substance is represented by the formula:
$Density(d) = \dfrac{{mass(m)}}{{Volume(V)}}$
The number of moles are represented by the formula:
$No.\,of\,moles\,(n) = \dfrac{{Given\,mass(m)}}{{Molar\,mass}}$
The mole fraction of a substance is represented by the formula:
$Mole\,fraction\,of\,component\,X = \dfrac{{Total\,no.\,of\,moles\,X}}{{Total\,no.\,of\,moles\,X + Total\,no.\,of\,moles\,Y}}$:

Complete step by step answer:
We are given that the solution is $20\%$ by mass. It means that $20g$ $KI$ is dissolved in $80g$ water. The mass of solute is$20g$ and the mass of solvent is $80g$ The density of the $KI$ solution is $1.202{{ }}g/mL$. Now we will calculate the number of moles of solute i.e. $KI$ by dividing the given mass with the molar mass.
The molar mass of $KI$ $= 39 + 127 = 166g$
Number of moles of $KI$ are given as:
$No.\,of\,moles\,(n) = \dfrac{{Given\,mass(m)}}{{Molar\,mass}}$
$No.\,of\,moles\,of\,KI\,(n) = \dfrac{{20}}{{166}} = 0.12$
Now we will calculate the volume of solution with the help of density.
Given, $d = 1.202{{ }}g/mL$ , $m = 20g$
Density is given as: $Density(d) = \dfrac{{mass(m)}}{{Volume(V)}}$
$1.202{{ }} = \dfrac{{100}}{{Volume(V)}}$
$Volume(V){{ }} = \dfrac{{100}}{{1.202}} = 83.19mL = 83 \times {10^{ - 3}}L$
Now we will calculate the molarity using the values of moles and volume. It will be:
$M = \dfrac{{No.\,of\,moles\,\,of\,solute}}{{Volume\,of\,solution(in\,L)}}$
$M = \dfrac{{0.12}}{{83.19 \times {{10}^{ - 3}}}} = 1.45$
Now, we will calculate the molarity using the values of moles and solvent. It will be:
$m = \dfrac{{No.\,of\,moles\,\,of\,solute}}{{mass\,of\,solvent(in\,Kg)}}$
$m = \dfrac{{0.12}}{{80 \times {{10}^{ - 3}}}} = 1.5$
Now, we will calculate the mole fraction of $KI$ with the help of moles of $KI$ and moles of water.
Total number of moles of water will be:
$No.\,of\,moles\,(n) = \dfrac{{Given\,mass(m)}}{{Molar\,mass}}$
$No.\,of\,moles\,of\,water\,(n) = \dfrac{{80}}{{18}} = 4.44$
The mole fraction of potassium iodide will be;
$Mole\,fraction\,of\,KI = \dfrac{{Total\,no.\,of\,moles\,of\,KI}}{{Total\,no.\,of\,moles\,of\,KI + Total\,no.\,of\,moles\,of\,water}}$
$Mole\,fraction\,of\,KI = \dfrac{{0.12}}{{0.12 + 4.44}} = \dfrac{{0.12}}{{4.56}} = 0.026$
Hence the molarity, molality and mole fraction of $20\%$ (mass/mass) aqueous $KI$ solution is $1.45,1.51$ and $0.026$ respectively.

Note:
The given compound is $KI$ i.e. potassium iodide. It is a metal halide salt of potassium and iodide ions. Molarity is defined as the number of moles of solute that are dissolved in $1$ litre of solution. It is denoted by $M$.