What mass of na2co3 is present in 0.650 l of a 0.505 m na2co3 solution?

V=0.650 L
c=0.505 mol/L
M(Na₂CO₃)=106.0 g/mol

the amount of substance of sodium carbonate is:
n(Na₂CO₃)=vc

the mass of sodium carbonate is:
m(Na₂CO₃)=n(Na₂CO₃)M(Na₂CO₃)

m(Na₂CO₃)=vcM(Na₂CO₃)

m(Na₂CO₃)=0.650*0.505*106.0=34.7945 g

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