Answer: The molarity of given solution is 0.0235 M.
Explanation:
Given : Volume of solution = 511 mL
Convert mL into L as follows.
[tex]1 mL = 0.001 L\\511 mL = 511 mL \times \frac{0.001 L}{1 mL}\\= 0.511 L[/tex]
Mass of [tex]NH_{3}[/tex] (solute) = 205 mg
Convert mg into gram as follows.
[tex]1 mg = 0.001 g\\205 mg = 205 mg \times \frac{0.001 g}{1 mg}\\= 0.205 g[/tex]
As molar mass of [tex]NH_{3}[/tex] is 17 g/mol. Hence, number of moles of [tex]NH_{3}[/tex] are calculated as follows.
[tex]No. of moles = \frac{mass given}{molar mass}\\= \frac{0.205 g}{17 g/mol}\\= 0.012 mol[/tex]
Molarity is the number of moles of a solute dissolved in a liter of solution.
[tex]Molarity = \frac{no. of moles}{Volume (in L)}[/tex]
Substitute the values into above formula as follows.
[tex]Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.012 mol}{0.511 L}\\= 0.0235 M[/tex]
Thus, we can conclude that the molarity of given solution is 0.0235 M.
