How many moles of ions will be present in 791.0 mL of 0.200 M solution of K3PO4?
Answers
Given : volume of solution of K₃PO₄ is 791 ml and concentration of K₃PO₄ in solution is 0.2 M.
To find : The number of moles of ions will be present in 791 ml of 0.2 M solution of K₃PO₄.
solution : number of moles of K₃PO₄ = volume of solution in L × concentration of K₃PO₄
= 791/1000 × 0.2
= 0.1582 moles
as after dissociation of K₃PO₄, three K⁺ and one PO₄¯ ions are formed.
so, total number of ions in one molecule of K₃PO₄ is 4.
so number of moles of ions = 4 × 0.1582 = 0.6328 mol
Therefore number of moles of ions will be present in 791 ml of 0.2M solution of K₃PO₄.
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