How many grams of Bacl2.2 H2O (244 g mol-1) should be taken to prepare 500 ml of 0.074 M CI- solution

Dear student,

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BaCl2 >> Ba^2+ + 2 Cl^- So, first note that for each mole of BaCl2 you get TWO moles of Cl. You want 500 ml of 0.074 M Cl, so 0.5L x 0.074 mol/L = 0.037 moles of Cl needed This will come from 1/2 that amount of solid BaCl2 or 0.0185 moles of BaCl2. 0.0185 moles x 244.3 g/mol = 4.52 grams of BaCl2
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