Answer:
Final concentration of NaOH = 0.75 M
Explanation:
For [tex]NaOH[/tex] :-
Given mass = 90.0 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{90.0\ g}{39.997\ g/mol}[/tex]
[tex]Moles\ of\ NaOH= 2.2502\ mol[/tex]
Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.
The expression for the molarity, according to its definition is shown below as:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Where, Volume must be in Liter.
It is denoted by M.
Given, Volume = 3.00 L
So,
[tex]Molarity=\frac{2.2502\ mol}{3.00\ L}=0.75\ M[/tex]
Final concentration of NaOH = 0.75 M
