Answer: The molar concentration of 29 g of [tex]Mg(OH)_{2}[/tex] dissolved in 1.00 L of water is 0.497 M.
Explanation:
Given: Mass = 29 g
Volume = 1.00 L
Moles is the mass of substance divided by its molar mass. So, moles of [tex]Mg(OH)_{2}[/tex] is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{29 g}{58.32 g/mol}\\= 0.497 mol[/tex]
Molarity is the number of moles of a substance divided by volume in liter.
Hence, molarity of the given solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.497 mol}{1.00 L}\\= 0.497 M[/tex]
Thus, we can conclude that the molar concentration of 29 g of [tex]Mg(OH)_{2}[/tex] dissolved in 1.00 L of water is 0.497 M.
