Question

Please help

Answer 1

Answer:  There are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.

Explanation:

Given: Molarity of solution = 2.0 M

Volume of solution = 250 mL

Convert mL int L as follows.

$1 mL = 0.001 L\\250 mL = 250 mL \times \frac{0.001 L}{1 mL}\\= 0.25 L$

Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given $BaSO_{4}$ solution is as follows.

$Molarity = \frac{mass}{Volume (in L)}\\2.0 M = \frac{mass}{0.25 L}\\mass = 0.5 g$

Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M $BaSO_{4}$ solution.