Question

What is the molarity of a solution made from 325.4 g of AlCl3 (MM 133.34 g/mol) to make 500.0 mL?

Answer 1

Answer : The molarity of a solution is, 4.88 M

Explanation : Given,

Mass of $AlCl_3$ = 325.4 g

Volume of solution = 500.0 mL

Molar mass of $AlCl_3$ = 133.34 g/mole

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

Formula used :

$\text{Molarity}=\frac{\text{Mass of }AlCl_3\times 1000}{\text{Molar mass of }AlCl_3\times \text{Volume of solution (in mL)}}$

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

$\text{Molarity}=\frac{325.4g\times 1000}{133.34g/mole\times 500.0mL}=4.88mole/L=4.88M$

Therefore, the molarity of a solution is, 4.88 M

Answer 2

Answer: The molarity of solution is 4.880 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

mass of $AlCl_3$ = $\frac{\text {given mass}}{\text {Molar Mass}}=\frac{325.4g}{133.34g/mol}=2.440$

Now put all the given values in the formula of molality, we get

$Molality=\frac{2.440\times 1000}{500ml}=4.880mole/L$

Therefore, the molarity of solution is 4.880 M