Question

Calculate the volume in mL of a 3.25 M solution of Al2(SO4)3 needed to provide 110 g of solute. ______ mL

Answer 1

Approximately
`92.9\; \rm mL`.

Step-by-step explanation:

Look up relevant relative atomic mass data on a modern periodic table:

• 
  `\rm Al`:
  `26.982`.
• 
  `\rm S`:
  `32.06`.
• 
  `\rm O`:
  `15.999`.

Calculate the formula mass of
`\rm Al_2(SO_4)_3`:

`M(\mathrm{Al_2(SO_4)_3}) = 2 * 26.982 + 3 * (32.06 + 4 * 15.999) = 342.132\; \rm g \cdot mol^(-1)`.

Calculate the number of moles of
`\rm Al_2(SO_4)_3` formula units that corresponds to
`110\; \rm g`:

`\displaystyle n(\mathrm{Al_2(SO_4)_3}) = (m)/(M) = (110\; \rm g)/(342.132\; \rm g \cdot mol^(-1)) \approx 0.322\; \rm mol`.

Calculate the volume of a
`3.25\; \rm M`
`\rm Al_2(SO_4)_3` solution that would contain
`0.322\; \rm mol` of
`\rm Al_2(SO_4)_3\!` formula units. Note, that the unit of concentration "
`\rm M`" is equivalent to
`\rm mol \cdot L^(-1)`.

`\displaystyle V = (n)/(c) = (0.322\; \rm mol)/(3.25\; \rm mol \cdot L^(-1)) \approx 0.0989\; \rm L`.

Convert that to milliliters as requested:

`\displaystyle V \approx 0.0989\; \rm L * (1000\; \rm mL)/(1\; \rm L) = 98.9\; \rm mL`.