Answer:
Approximately [tex]3.06 \times 10^{2}\; \rm g[/tex] (approximately [tex]306\; \rm g[/tex].)
Explanation:
Calculate the quantity [tex]n[/tex] of lithium phosphate in [tex]V = 2.5\; \rm L[/tex] of this[tex]c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}[/tex] lithium phosphate solution.
[tex]\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}[/tex].
Empirical formula of lithium phosphate: [tex]\rm Li_3PO_4[/tex].
Look up the relative atomic mass of [tex]\rm Li[/tex], [tex]\rm P[/tex],and [tex]\rm O[/tex] on a modern periodic table:
Calculate the formula mass of [tex]\rm Li_3PO_4[/tex]:
[tex]M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}[/tex].
Calculate the mass of that [tex]n = 2.65\; \rm mol[/tex] of [tex]\rm Li_3PO_4[/tex] formula units:
[tex]\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}[/tex].
