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[tex]\text{Molarity,}~ S= 2.1~mol/L\\\\\text{Molar mas of}~ NaHCO_3, ~M=84~~ g/mol\\\\\text{Volume,}~~ V = 1 ~L\\\\\text{Mass,} ~w = ?[/tex]
[tex]\text{ We know that,}\\\\~~~~~S = \dfrac{n}V\\\\\implies n = SV\\\\\implies \dfrac{w}{M} = SV\\\\\implies w = SVM\\\\\implies w= 2.1 \times 1 \times 84\\\\\implies w =176.4 ~ g[/tex]