Question

When $1.04g$ of $Ba{Cl}_2$ is present in ${10}^{5}g$ of solution, the concentration of solution is:

• A. $0.104ppm$
• B. $10.4ppm$
• C. $0.0104ppm$
• D. $104ppm$

Answer 1

Answer: (B)

$10.4ppm$

Concentration in ppm $=\frac{\text{mass of solute}}{\text{mass of solution}}\times {10}^6$

$=\frac{\text{mass of}BaC{l}_2}{\text{mass of solution}}\times {10}^6$

$=\frac{1.04g}{{10}^{5}g}\times {10}^6$

$=10.4ppm$

Hence option B is correct.