Question

An anhydrous (water remove) salt has a formula mass of 186.181 g/mol. If the hydrated version of the salt has 8 mol of water associated with it, what is the mass % of water in the hydrated salt?

Answer 1

Answer: The mass percent of water in the hydrated salt is 43.6 %

Explanation:

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 8 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$8moles=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(8mol\times 18g/mol)=144g$

We are given:

Mass of anhydrous salt = 186.181 g

To calculate the mass percentage of water in the hydrated salt, we use the equation:

$\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Mass of hydrated salt}}\times 100$

Mass of hydrated salt = [186.181 + 144]g = 330.181g

Mass of water = 144 g

Putting values in above equation, we get:

$\text{Mass percent of water}=\frac{144g}{330.181g}\times 100=43.6\%$

Hence, the mass percent of water in the hydrated salt is 43.6 %

Answer 2

Answer:

The mass percent of water in the hydrated salt is 43.6 %Explanation: