Question

Please help

How many grams of barium sulfate are present in 250 of 2.0 M BaS04 solution?

Answer 1

Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M
`BaSO_(4)` solution.

Step-by-step 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 * (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 = (mass)/(Volume (in L))\\2.0 M = (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.