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Answer to Question #27247 in Inorganic Chemistry for sali

Question #27247
what weight of sodium carbonate must be taken to give 100 cm^3 of a 0.1 M solution?
Expert's answer
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 MW(gmol)
Themolar weight of Na2CO3
MW(Na2CO3) = 2 x MW(Na) + MW(C) + 3 x MW(O)

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

Answer: n(Na2CO3) = 1.06 g

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