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
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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