Question

What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium

Answer 1

Answer:

Approximately $3.06 \times 10^{2}\; \rm g$ (approximately $306\; \rm g$.)

Explanation:

Calculate the quantity $n$ of lithium phosphate in $V = 2.5\; \rm L$ of this$c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ lithium phosphate solution.

$\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}$.

Empirical formula of lithium phosphate: $\rm Li_3PO_4$.

Look up the relative atomic mass of $\rm Li$, $\rm P$,and $\rm O$ on a modern periodic table:

• $\rm Li$: $6.94$.
• $\rm P$: $30.974$.
• $\rm O$: $15.999$.

Calculate the formula mass of $\rm Li_3PO_4$:

$M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}$.

Calculate the mass of that $n = 2.65\; \rm mol$ of $\rm Li_3PO_4$ formula units:

$\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}$.