Question

Question 8 4 pts What would be the resulting molarity of a solution made by dissolving 31.3 grams of Ca(OH)2 in enough water to make a 1050-milliliter solution? Show all of the work needed to solve this problem.

Answer 1

Answer : The molarity of the solution is, 0.4028 mole/L

Explanation : Given,

Mass of $Ca(OH)_2$ = 31.3 g

Molar mass of $Ca(OH)_2$ = 74 g/mole

Volume of solution = 1050 ml

Molarity : It is defined as the moles of solute present in one liter of solution.

Formula used :

$Molarity=\frac{\text{Mass of }Ca(OH)_2\times 1000}{\text{Molar mass of }Ca(OH)_2\times \text{volume of solution in ml}}$

Now put all the given values in this formula, we get:

$Molarity=\frac{31.3g\times 1000}{74g/mole\times 1050ml}=0.4028mole/L$

Therefore, the molarity of the solution is, 0.4028 mole/L

Answer 2

Answer is: molarity of a solution is 0,401 M.
m(Ca(OH)₂) = 31,3 g.
n(Ca(OH)₂) = m(Ca(OH)₂) ÷ M(Ca(OH)₂).
n(Ca(OH)₂) = 31,3 g ÷ 74 g/mol.
n(Ca(OH)₂) = 0,422 mol.
V(solution) = 1050 mL · 0,001 L/mL = 1,050 L.
c(Ca(OH)₂) = n(Ca(OH)₂) ÷ V(solution).
c(Ca(OH)₂) = 0,422 mol ÷ 1,050 L.
n(Ca(OH)₂) = 0,401 mol/L = 0,401 M.