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.

Step-by-step 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 = (mass)/(molar mass)\\= (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 = (moles)/(Volume (in L))\\= (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.