10) a 3.34 g sample of a hydrate has the formula srs₂o₃·xh₂o, and contains 2.30 g of the anhydrous…

Explanation:

Step1: Calculate mass of water

The mass of water in the hydrate is the mass of the hydrate minus the mass of the anhydrous salt. So, $m_{H_2O}=3.34\ g - 2.30\ g=1.04\ g$.

Step2: Calculate moles of anhydrous salt

The molar - mass of $SrS_2O_3$ is $M_{SrS_2O_3}=87.62\ g/mol+2\times32.07\ g/mol + 3\times16.00\ g/mol=87.62\ g/mol + 64.14\ g/mol+48.00\ g/mol = 199.76\ g/mol$. The moles of $SrS_2O_3$, $n_{SrS_2O_3}=\frac{m_{SrS_2O_3}}{M_{SrS_2O_3}}=\frac{2.30\ g}{199.76\ g/mol}\approx0.0115\ mol$.

Step3: Calculate moles of water

The molar - mass of $H_2O$ is $M_{H_2O}=2\times1.01\ g/mol + 16.00\ g/mol = 18.02\ g/mol$. The moles of $H_2O$, $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{1.04\ g}{18.02\ g/mol}\approx0.0577\ mol$.

Step4: Find the ratio of moles of water to moles of anhydrous salt

The ratio $x=\frac{n_{H_2O}}{n_{SrS_2O_3}}=\frac{0.0577\ mol}{0.0115\ mol}\approx5$.

Answer:

5