what is the percent by mass of water in na₂so₄·10h₂o? use %element = \\frac{total mass of element in…

Explanation:

Step1: Calculate molar mass of water

The molar mass of $H_2O$ is $2\times1.01 + 16.00=18.02\ g/mol$. In $Na_2SO_4\cdot10H_2O$, there are 10 moles of $H_2O$, so the total mass of water in the compound is $10\times18.02 = 180.2\ g/mol$.

Step2: Calculate molar mass of $Na_2SO_4\cdot10H_2O$

The molar mass of $Na$ is $22.99\ g/mol$, $S$ is $32.07\ g/mol$, $O$ is $16.00\ g/mol$. For $Na_2SO_4$, it is $2\times22.99+32.07 + 4\times16.00=142.05\ g/mol$. Adding the mass of 10 moles of water ($180.2\ g/mol$), the molar mass of $Na_2SO_4\cdot10H_2O$ is $142.05+180.2 = 322.25\ g/mol$.

Step3: Calculate percent - by - mass of water

Using the formula $%\text{Element}=\frac{\text{total mass of element in compound}}{\text{molar mass of compound}}\times100$, we have $\frac{180.2}{322.25}\times 100\approx55.92%$.

Answer:

55.92%