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A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What is the molarity of this solution? Express your answer to four significant figures and include the appropriate units.

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Answer:

Approximately
1.854\; \rm mol\cdot L^(-1).

Step-by-step explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

Formula mass of strontium hydroxide

Look up the relative atomic mass of
\rm Sr,
\rm O, and
\rm H on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.


  • \rm Sr:
    87.62.

  • \rm O:
    15.999.

  • \rm H:
    1.008.

Calculate the formula mass of
\rm Sr(OH)_2:


M\left(\rm Sr(OH)_2\right) = 87.62 + 2* (15.999 + 1.008) = 121.634\; \rm g \cdot mol^(-1).

Number of moles of strontium hydroxide in the solution


M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^(-1) means that each mole of
\rm Sr(OH)_2 formula units have a mass of
121.634\; \rm g.

The question states that there are
10.60\; \rm g of
\rm Sr(OH)_2 in this solution.

How many moles of
\rm Sr(OH)_2 formula units would that be?


\begin{aligned}n\left(\rm Sr(OH)_2\right) &= (m\left(\rm Sr(OH)_2\right))/(M\left(\rm Sr(OH)_2\right))\\ &= (10.60\; \rm g)/(121.634\; \rm g \cdot mol^(-1)) \approx 8.71467* 10^(-2)\; \rm mol\end{aligned}.

Molarity of this strontium hydroxide solution

There are
8.71467* 10^(-2)\; \rm mol of
\rm Sr(OH)_2 formula units in this
47\; \rm mL solution. Convert the unit of volume to liter:


V = 47\; \rm mL = 0.047\; \rm L.

The molarity of a solution measures its molar concentration. For this solution:


\begin{aligned}c\left(\rm Sr(OH)_2\right) &= (n\left(\rm Sr(OH)_2\right))/(V)\\ &= (8.71467* 10^(-2)\; \rm mol)/(0.047\; \rm L) \approx 1.854\; \rm mol \cdot L^(-1)\end{aligned}.

(Rounded to four significant figures.)

User Augustino
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