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

Answer:

Approximately $92.9\; \rm mL$.

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 $\rm Al_2(SO_4)_3$:

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

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

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

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 $\rm mol \cdot L^{-1}$.

$\displaystyle V = \frac{n}{c} = \frac{0.322\; \rm mol}{3.25\; \rm mol \cdot L^{-1}} \approx 0.0989\; \rm L$.

Convert that to milliliters as requested:

$\displaystyle V \approx 0.0989\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 98.9\; \rm mL$.