When 1.04 g of BaCl_2 is dissolved in 10^5 | vedclass

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Question

(A) The formula for concentration in parts per million $(ppm)$ is given by: $ppm = \frac{	ext{mass of solute}}{	ext{mass of solution}} 	imes 10^6$

Vedclass · Accepted Answer

$10.4$

Vedclass · Answer

(A) The formula for concentration in parts per million $(ppm)$ is given by: $ppm = \frac{	ext{mass of solute}}{	ext{mass of solution}} 	imes 10^6$ Given: Mass of solute $(BaCl_2)$ = $1.04 \ g$ Mass of solvent (water) = $10^5 \ g$ Since the mass of solute is very small compared to the mass of solvent,the mass of the solution is approximately equal to the mass of the solvent: Mass of solution $\approx 10^5 \ g$ Now,substituting the values: $ppm = \frac{1.04}{10^5} 	imes 10^6$ $ppm = 1.04 	imes 10^1 = 10.4$ Therefore,the concentration is $10.4 \ ppm$.

When $1.04 \ g$ of $BaCl_2$ is dissolved in $10^5 \ g$ of water,what will be the concentration of the solution in $ppm$?

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