Molar Concentration (C) =0.250 mol/LVolume (V) =150mL=0.150Lnumber of moles (n) =C×V=0.250×0.150=0.0375moles\begin{aligned} \textsf{Molar Concentration (C) }&= 0.250\ mol/L\\ \textsf{Volume (V) } &= 150mL = 0.150L\\ \\ \textsf{number of moles (n) }&= C × V\\ &= 0.250 × 0.150\\ &= 0.0375 moles \end{aligned}Molar Concentration (C) Volume (V) number of moles (n) =0.250 mol/L=150mL=0.150L=C×V=0.250×0.150=0.0375moles
mass of solute =number of moles ×molar massmolar mass of Fe(NO3)3=56+3(14+48)=242 g/mol∴mass of solute =0.0375×242=9.075g\begin{aligned} \textsf{mass of solute } &= \textsf{number of moles } × \textsf{molar mass}\\ \\ \textsf{molar mass of }Fe({NO_3})_3 &= 56 + 3(14 + 48) = 242\ g/mol\\ \\ \therefore \textsf{mass of solute } &= 0.0375 × 242\\ &= 9.075g \end{aligned}mass of solute molar mass of Fe(NO3)3∴mass of solute =number of moles ×molar mass=56+3(14+48)=242 g/mol=0.0375×242=9.075g
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