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Answer: The molarity of 8 grams of an aqueous solution of sodium hydroxide in 2 liters of solute is 0.1 M
Explanation:
Given: Mass of solute = 8 g
Volume of solution = 2 L
Molar mass of NaOH is 40 g/mol.
Number of moles of NaOH are calculated as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{8 g}{40 g/mol}\\= 0.2 mol[/tex]
As molarity is the number of moles of solute present in a liter of solution. Therefore, molarity of given solution is calculated as follows.
[tex]Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.2 mol}{2 L}\\= 0.1 M[/tex]
Thus, we can conclude that the molarity of 8 grams of an aqueous solution of sodium hydroxide in 2 liters of solute is 0.1 M.
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