Question
Question: When 1.04g of \(BaC{l_2}\) is present in \({10^5} g\) of solution, the concentration of solution is:...
When 1.04g of BaCl2 is present in 105g of solution, the concentration of solution is:
(A) 0.104 ppm
(B) 10.4 ppm
(C) 0.0104 ppm
(D) 1.4 ppm
Solution
If any homogeneous mixture is given to us which is of binary type that is one solute and solvent, we find out the concentration of solution by using various methods. These methods involve some physical (percentage concentration, ppm, ppb, etc.) and some chemical (mole fraction, molarity, molality, normality, etc.) The given method is based on a simple calculation of concentration where we consider the amount of solute and solvent.
Formula used: Parts per million (ppm) =Mass of solutionMass of solute×106
Complete answer:
From the given formula
Parts per million (ppm) =Mass of solutionMass of solute×106
Here, solute is barium chloride and solution is mixture of barium chloride in water
Now, Concentration of solution in ppm=weight of solutionweight of BaCl2×106
Weight of solute (BaCl2)=1.04g
Weight of solution=105g
Now putting the values of barium chloride and weight of solution, we get
Concentration of solution in ppm =weight of solutionweight of solute(BaCl2)×106
∴ Concentration of solution in ppm
=105g1.04g×106
Solving the equation, we get
The concentration of solution in ppm =10.40ppm
Hence the correct answer is option (B).
Note: Parts per million (ppm) is defined as the number of parts by mass of solute per million parts of solution. i.e. 10.40ppm of BaCl2 dissolved in 106g of solution.