Given, the required #M# and the needed volume for the solution, the mass of the sodium chloride #(NaCl)# can be computed as;
#"Molarity"(M)=("no. of moles"(eta))/("Li Solution"(V))#
where:
#eta=(mass(m))/("Molar mass"(Mm)#
Now find the molar mass of #NaCl# . Value is obtainable from the periodic table; i.e.,
#NaCl=(58.5g)/(mol)#
Substitute the value of #eta# to the original formula. Rearrange it to isolate the #m# and plug in given values to solve the required data as shown below;
#M=(m/(Mm))/V#
#(m)/(Mm)=MV#
#m=MxxVxxMm#
#m=(3.5cancel(mol))/(cancel(L))xx(1cancel(L))xx(58.5g)/(1cancel(mol))#
#m=204.75g#