Question

What is the molarity of a solution that is made by mixing 35.5 g of Ba(OH)2 in 325 ml of solution?

Answer 1

`M=0.638M`

Step-by-step explanation:

Hello!

In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:

`n=35.5g Ba(OH)_2*(1molBa(OH)_2)/(171.34gBa(OH)_2)\\\\n=0.207mol`

Then, the liters of solution:

`V=325mL*(1L)/(1000mL) =0.325L`

Finally, the molarity turns out:

`M=(0.207mol)/(0.325L)\\\\M=0.638M`

Best regards!