Answer: The molarity of 8 grams of an aqueous solution of sodium hydroxide in 2 liters of solute is 0.1 M
Step-by-step 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.
![No. of moles = (mass)/(molar mass)\\= (8 g)/(40 g/mol)\\= 0.2 mol](https://img.qammunity.org/2022/formulas/chemistry/high-school/8lybllcy05t26s7o4mm00lf62q88v8d3te.png)
As molarity is the number of moles of solute present in a liter of solution. Therefore, molarity of given solution is calculated as follows.
![Molarity = (no. of moles)/(Volume (in L))\\= (0.2 mol)/(2 L)\\= 0.1 M](https://img.qammunity.org/2022/formulas/chemistry/high-school/9kxp69e60blrxg841x1ax5qopbp9gga1ns.png)
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.