Question

calculate the no. of molecules of hydration and no. of chlorine atoms in 107g of barium chlorate Ba(ClO3)2 . H2O​

Answer 1

Explanation:

Calculate the number of molecules of hydration and number of Cl atoms in 107.0 g of barium chlorate At mass of Ba is 137u. Dear student, The mole of the Ba(ClO3)2·H2O) is calculated as follows, Mole of the Ba(ClO3)2·H2O) = 107 g / 322.2460 g/mole = 0.33204 mole.

Answer 2

Explanation:

The mole of the Ba(ClO3)2·H2O) is calculated as follows,

Mole of the Ba(ClO3)2·H2O) = 107 g / 322.2460 g/mole = 0.33204 mole.

The number of the hydration molecule is calculated as follows,

Number of hydration molecule = (0.33204 mole Ba(ClO3)2·H2O) x (1 mol H2O / 1 Ba(ClO3)2·H2O) x (6.023 x 10^23 molecules/mol)

Number of hydration molecule = 2.00 x 10^23 molecules

The number of the chlorine atom is calculated as follows,

Number of the chlorine atom = (0.33204 mol Ba(ClO3)2·H2O) x (2 mol Cl / 1 Ba(ClO3)2·H2O) x (6.023 x 10^23 atoms/mol)

Number of the chlorine atom= 4.00 x 10^23 atoms

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