Question

What is the molar concentration of 29 g of Mg(OH)2 dissolved in 1.00 L of water

Answer 1

Answer: The molar concentration of 29 g of $Mg(OH)_{2}$ 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 $Mg(OH)_{2}$ is as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{29 g}{58.32 g/mol}\\= 0.497 mol$

Molarity is the number of moles of a substance divided by volume in liter.

Hence, molarity of the given solution is as follows.

$Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.497 mol}{1.00 L}\\= 0.497 M$

Thus, we can conclude that the molar concentration of 29 g of $Mg(OH)_{2}$ dissolved in 1.00 L of water is 0.497 M.