Answer to Question #27247 in Inorganic Chemistry for sali

Question

what weight of sodium carbonate must be taken to give 100 cm^3 of a 0.1 M solution?

Answer 1

Theamount of Na₂CO₃ is
n(Na₂CO₃) = C(M) x V(L) = 0.1 x 0.1 = 0.01mol
C -concentration of Na₂CO₃ solution (M)
V - volume of solution (L) [ 100cm^3 = 100 mL = 0.100L ]
The mass of Na₂CO₃ is

m(g) = n(mol) x M_W(gmol)
Themolar weight of Na₂CO₃
M_W(Na₂CO₃) = 2 x M_W(Na) + M_W(C) + 3 x M_W(O)

m(Na₂CO₃) = 0.01 x (23 x 2 +12+48) = 1.06 g

Answer: n(Na₂CO₃) = 1.06 g