A 100 mL solution was made by adding 1 . 43 g of Na 2 CO 3 . xH 2 O . The normality of the solution is 0 . 1 N . The value of x is _______ (The atomic mass of Na is 23 g / mol )

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Equivalent of solute = 0 . 1 × 0 . 1 Mole of solute Na 2 CO 3 , xH 2 O = [ 0 . 1 × 0 . 1 ] 1 2 Mass of Na 2 CO 3 . xH 2 O = [ 0 . 1 × 0 . 1 ] 1 2 × [ 106 + 18 x ] = 1 . 43 ⇒ [ 106 + 18 x = 286 18 x = 180 x = 10

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