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Question: A \(1.24M\) aqueous solution of \(KI\) has density of \(1.15g/c{m^3}\) . What is the percentage comp...

A 1.24M1.24M aqueous solution of KIKI has density of 1.15g/cm31.15g/c{m^3} . What is the percentage composition of solute in the solution?
A. 17.8917.89
B. 27.8927.89
C. 37.8937.89
D. 47.8947.89

Explanation

Solution

In this question we have to calculate weight by weight percentage composition. We will calculate the weight of the solute and the solution and get their ratio multiplied by 100100 to get the answer. Also molar mass of KIKI is 166g/mol166g/mol .
Formula used:
M=(nb/V)×1000M = ({n_b}/V) \times 1000
Where, MM is the molarity
nb{n_b} is the number of moles of solute
And VV is the volume of the solution in mlml or cm3c{m^3}
Percentage composition of solute in solution
%(w/w)=(Wb/Wsolution)×100\% (w/w) = ({W_b}/{W_{solution}}) \times 100
w=w = weight
Wb={W_b} = weight of solute
Wsolution={W_{solution}} = weight of solution

Complete step by step answer:
In the question, the following information and values are given:
Molarity is given to be 1.24M1.24M
Since we know that molarity is the number of moles of solute present in one litre of the solution , we can say that
1l1l of the solution ( 1000ml1000ml ) has 1.24mol1.24mol of KIKI
Hence, 1l1l of the this solution will have 1.24×166g1.24 \times 166g of KIKI =205.84g = 205.84g (this is the weight of the solute or Wb{W_b} )
[we know that, number of moles=Givenweightmolecularweight\text{number of moles} = \dfrac{Given weight}{molecular weight}
Hence, given weight of KIKI =1.24×166g = 1.24 \times 166g ]
Density of the solution is given to be 1.15g/cm31.15g/c{m^3}
This means that in 1cm31c{m^3} volume, we have 1.15g1.15g
Thus in 1000cm31000c{m^3} we will have 1.15×1000g1.15 \times 1000g
Or we can say that 1l1l has 1150g1150g (weight of the solution or Wsolution{W_{solution}} )
Thus, now we will use the percentage composition formula
%(w/w)=(Wb/Wsolution)×100\% (w/w) = ({W_b}/{W_{solution}}) \times 100
Putting all the values, we will have
%(w/w)=(205.8/1150)×100 %(w/w)=17.89%  \% (w/w) = (205.8/1150) \times 100 \\\ \Rightarrow \% (w/w) = 17.89\% \\\
Thus the answer is Option A .

Additional Information: Molarity is defined as the number of moles of solute which are present in one litre of a solution. It is a concentration term, other concentration terms could be normality which is the number of gram equivalent of a solute present in one litre of a solution. Molality is also a widely used concentration term which means number of moles of a solute present in one kilogram of a solvent.

Note:
Percentage composition means the parts of solute which are there per 100 parts of the solution ( mass of solution is a sum of the mass of the solute and the mass of the solvent).