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Q.
When $1.04\, g$ of $BaCl_2$ is present in $10^{5} \,g$ of solution the concentration of solution is
Solutions
Solution:
Concentration in ppm $=\frac{\text { mass of solute }}{\text { mass of solution }} \times 10^{6}$
$
=\frac{\text { mass of } BaCl _{2}}{\text { mass of solution }} \times 10^{6}
$
$
\begin{array}{l}
=\frac{1.04 g }{10^{5} g } \times 10^{6} \\
=10.4 ppm
\end{array}
$