Answer :
Answer:
The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g
Explanation:
We do this by preparing the equation:
Mass = concentration (mol/L) x volume (L) x Molar mass
Mass = 1.0 M x 2.5 L x 40 g/mol
Mass = 100 g
Answer:
[tex]100\; \rm g[/tex] of [tex]{\rm NaOH}\; (s)[/tex] would be required.
Explanation:
The quantity of solute in a solution of concentration [tex]c[/tex] and volume [tex]V[/tex] would be [tex]n = c \cdot V[/tex].
It is given that volume [tex]V = 2.5\; \rm L[/tex] for the solution in this question. It is also given that the concentration of the [tex]\rm NaOH[/tex] solute in this solution is [tex]c = 1.0\; \rm M[/tex], which is the equivalent to [tex]c = 1.0\; \rm mol \cdot L^{-1}[/tex].
Apply the equation [tex]n = c \cdot V[/tex] to find the quantity of [tex]\rm NaOH[/tex] in this solution:
[tex]\begin{aligned}n &= c \cdot V \\ &= 1.0\; \rm mol \cdot L^{-1} \times 2.5\; \rm L \\ &= 2.5\; \rm mol\end{aligned}[/tex].
Multiply the quantity [tex]n[/tex] of [tex]\rm NaOH[/tex] in this solution with the formula mass [tex]M[/tex] of [tex]{\rm NaOH}\![/tex] to find the corresponding mass:
[tex]\begin{aligned}m &= n \cdot M \\ &= 2.5\; \rm mol \times 40\; \rm g \cdot mol^{-1} \\ &= 100\; \rm g\end{aligned}[/tex].
Thus, this solution would contain [tex]100\; \rm g[/tex] of [tex]{\rm NaOH}[/tex].
It would thus take [tex]100\; \rm g[/tex] of [tex]{\rm NaOH}[/tex] to prepare this solution.