The number of moles of NaCl required to form 5M of 450 ml solution is,
[tex]\begin{gathered} N=450ml\times\frac{5}{1000}Mml^{-1} \\ N=2.25M\text{ } \end{gathered}[/tex]The mass of the NaCl is,
[tex]M=m_{Na}+m_{Cl}[/tex]Substituting the known values,
[tex]\begin{gathered} M=22.99+35.453 \\ M=58.443\text{ gm} \end{gathered}[/tex]Thus, the amount of NaCl required in gm to make the given solution is,
[tex]\begin{gathered} M^{\prime}=M\times N \\ M^{\prime}=58.443\times2.25 \\ M^{\prime}=131.5\text{ gm} \end{gathered}[/tex]Hence, the amount of NaCl equired for the given solution is 131.5 gm
