Respuesta :
Answer:
The molarity of this [tex]\rm NaOH[/tex] solution is approximately [tex]13.0\; \rm mol \cdot L^{-1}[/tex].
Explanation:
The molarity of a substance gives its quantity (in number of moles) in each liter of the solution.
According to the question, there are [tex]521\; \rm g[/tex] of [tex]\rm NaOH[/tex] in each liter of this solution. The goal is to find the number of moles of formula units in that
[tex]\rm NaOH[/tex] contains three elements: Na, O, and H. Look up a modern periodic table for the relative atomic mass of each element:
- Na: 22.990.
- O: 15.999.
- H: 1.008.
Calculate the formula mass of [tex]\rm NaOH[/tex].
[tex]M(\rm NaOH) = 22.990 + 15.999 +1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].
That's the mass of each mole of [tex]\rm NaOH[/tex] formula units. Calculate the number of moles of formula units in that [tex]521\; \rm g[/tex] of
[tex]\displaystyle n = \frac{m}{M} = \frac{521\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 13.0\; \rm mol[/tex].
Apply the equation [tex]\displaystyle c = \frac{n}{V}[/tex] to find the molar concentration [tex]c[/tex] of this solution. In this equation, [tex]n[/tex] is the number of moles of the solute, and [tex]V[/tex] is the volume of the solution.
[tex]\begin{aligned}c &= \frac{n}{V} \\ &\approx \frac{13.0\; \rm mol}{1\; \rm L} = 13.0\; \rm mol\cdot L^{-1}\end{aligned}[/tex].