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2.65 g of Ba(OH)2 is dissolved in 70.0 mL of water to produce a saturated solution...

2.65 g of Ba(OH)2 is dissolved in 70.0 mL of water to produce a saturated solution at 200C. Calculate the solubility in units of g/100 mL; g/L ; and M.

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Expert Solution

Molar mass of Ba(OH)2 = [1*137.327 + 2*(15.9994 + 1.008)] g/mol = 171.3418 g/mol.

Moles of Ba(OH)2 = (2.65 g)/(171.3418 g/mol) = 0.01547 mole.

Volume of water taken = 70.0 mL = (70.0 mL)*(1 L/1000 mL) = 0.07 L.

70.0 mL water contains 2.65 g Ba(OH)2.

Therefore, 100.0 mL water contains (2.65 g)*(100 mL/70 mL) = 3.7857 g ≈ 3.786 g.

Solubility of Ba(OH)2 = 3.786 g/100 mL (ans).

Again 100 mL water = (100 mL)*(1 L/1000 mL) = 0.100 L.

Solubility of Ba(OH)2 in a liter of water = (3.786 g/100 mL) = (3.786 g/0.100 L) = 37.86 g/L (ans).

Solubility of Ba(OH)2 in M = (0.01547 mole)/[(70.0 mL)*(1 L/1000 mL)] = (0.01547 mole)/(0.07 L) = 0.221 mol/L = 0.221 M (ans).

queen_honey_blossom answered 2 years ago
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