Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3 × 16) = 106 g mol-1
Now, 1 mole of Na2CO3 means 106 g of Na2CO3.
∴ 0.5 mol of Na2CO3 = \(\frac {106\ g}{1 \ mole} × 0.5\ mol \) Na2CO3
= 53 g Na2CO3
⇒ 0.50 M of Na2CO3 = 0.50 mol/L Na2CO3
Hence, 0.50 mol of Na2CO3 is present in 1L of water or 53 g of Na2CO3 is present in 1L of water.
LIST I | LIST II | ||
A. | \(K_2SO_4(aq)\) with 60% dissociation | I. | \(i =3.7\) |
B. | \(K_3[Fe(CN)_6](aq)\) with 90% dissociation | II. | \(i= 1.8\) |
C. | \(AlCl_3 (aq)\) with 80% dissociation | III. | \(i=2.2\) |
D. | \(K_2HgI_4 (aq)\) with 40% dissociation | IV. | \(i= 3.4\) |