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

Answer:

$M=0.638M$

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:

$n=35.5g Ba(OH)_2*\frac{1molBa(OH)_2}{171.34gBa(OH)_2}\\\\n=0.207mol$

Then, the liters of solution:

$V=325mL*\frac{1L}{1000mL} =0.325L$

Finally, the molarity turns out:

$M=\frac{0.207mol}{0.325L}\\\\M=0.638M$

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