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.

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 [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

[tex]\rm Sr[/tex]: [tex]87.62[/tex].
[tex]\rm O[/tex]: [tex]15.999[/tex].
[tex]\rm H[/tex]: [tex]1.008[/tex].

Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:

[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].

Number of moles of strontium hydroxide in the solution

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

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

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

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

Molarity of this strontium hydroxide solution

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

[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].

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

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

(Rounded to four significant figures.)