Respuesta :

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

The mass of the Na₂CO₃ in given solution is  34.794 g .

Given Here,

Volume (v) = 0.650 L

Molarity (M)  = 0.505 mol/L

Molar mass (m) of Na₂CO₃ = 106.0 g/mol

The mass of the Na₂CO₃,

w = V x M x m

put the values in the formula,

w = (0.650) x (0.505) x (106.0)

w = 34.794 g

Therefore, the mass of the Na₂CO₃ in given solution is  34.794 g .

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