Question

When 1.04g of $BaC{l_2}$ is present in ${10^5} g$ of solution, the concentration of solution is:
(A) 0.104 ppm
(B) 10.4 ppm
(C) 0.0104 ppm
(D) 1.4 ppm

Answer 1

Hint: 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) $ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times {10^6}$

Complete answer:
From the given formula
Parts per million (ppm) $ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times {10^6}$
Here, solute is barium chloride and solution is mixture of barium chloride in water
Now, Concentration of solution in $ppm = \dfrac{{weight{\text{ of BaC}}{{\text{l}}_2}}}{{weight{\text{ of solution}}}} \times {10^6}$
Weight of solute $\left( {BaC{l_2}} \right) = 1.04g$
Weight of solution$ = {10^5}g$
Now putting the values of barium chloride and weight of solution, we get
Concentration of solution in ppm $ = \dfrac{{weight{\text{ of solute}}\left( {BaC{l_2}} \right)}}{{weight{\text{ of solution}}}} \times {10^6}$
$\therefore $ Concentration of solution in ppm
$ = \dfrac{{1.04g}}{{{{10}^5}g}} \times {10^6}$
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 $BaC{l_2}$ dissolved in ${10^6}g$ of solution.