Answer:
Approximately [tex]0.47\; \rm mol \cdot L^{-1}[/tex] (note that [tex]1\; \rm M = 1 \; \rm mol \cdot L^{-1}[/tex].)
Explanation:
The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this [tex]\rm NaNO_3[/tex] solution in water,
Let [tex]n[/tex] be the number of moles of the solute in the whole solution. Let [tex]V[/tex] represent the volume of that solution. The formula for the molarity [tex]c[/tex] of that solution is:
[tex]\displaystyle c = \frac{n}{V}[/tex].
In this question, the volume of the solution is known to be [tex]1250\; \rm mL[/tex]. That's [tex]1.250\; \rm L[/tex] in standard units. What needs to be found is [tex]n[/tex], the number of moles of [tex]\rm NaNO_3[/tex] in that solution.
The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of [tex]\rm NaNO_3[/tex] is [tex]85\; \rm g \cdot mol^{-1}[/tex] means that the mass of one mole of
[tex]\displaystyle n = \frac{m}{M}[/tex].
For this question,
[tex]\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}[/tex].
Calculate the molarity of this solution:
[tex]\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}[/tex].
Note that [tex]1\; \rm mol \cdot L^{-1}[/tex] (one mole per liter solution) is the same as [tex]1\; \rm M[/tex].
