Answer to Question #348237 in General Chemistry for cece

General Chemistry

Question #348237

How many grams of sodium bicarbonate, NaHCO₃, are needed to make 200.0 ml of a 0.250 M solution?

Expert's answer

n(NaHCO₃)=c×V=0.2×0.25=0.05 mol

M(NaHCO₃)=23+1+12+48=84 g/mol

m(NaHCO₃)=n×M=0.05×84=4.2 g

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