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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:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of water = 8 moles
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]8moles=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(8mol\times 18g/mol)=144g[/tex]
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:
[tex]\text{Mass percent of water}=\frac{\text{Mass of water}}{\text{Mass of hydrated salt}}\times 100[/tex]
Mass of hydrated salt = [186.181 + 144]g = 330.181g
Mass of water = 144 g
Putting values in above equation, we get:
[tex]\text{Mass percent of water}=\frac{144g}{330.181g}\times 100=43.6\%[/tex]
Hence, the mass percent of water in the hydrated salt is 43.6 %
The mass percent of water in the hydrated salt is 43.6 %
Firstly, find the mass of water using given number of moles.
[tex]\text{ Number of moles}=\frac{\Given Mass}{\text{Molar mass}}[/tex]
Moles of water = 8 moles
Molar mass of water = 18 g/mol
On substituting the values:
[tex]\text{Mass}= \text{Molar mass} * \text{ Number of moles}= 18 \text{g/mol} * 8 \text{moles} =144 \text{g}[/tex]
Given:
Mass of anhydrous salt = 186.181 g
In order to calculate the mass percentage of water in the hydrated salt, the formula to be used is:
[tex]\text{ Mass percent of water}= \frac{\text{Mass of water}}{\text{Mass of hydrated salt}} * 100[/tex]
Since, Hydrated salt= (water+ salt)
Therefore mass of hydrated salt= [186.181 g + 144 g]= 330.181 g
Substituting the values in above equation, we get:
[tex]\text{ Mass percent of water}= \frac{144}{330.181} * 100=43.6\%[/tex]
Hence, the mass percent of water in the hydrated salt is 43.6 %
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