How many moles of k+ are present in 343 ml of a 1.27 m solution of k3po4

1 M or molar solution is defined as the solution in which 1 mole of a compound or substance is dissolved in 1 litre of total volume. The 1M is equal to 1 mole of the compound present in 1 L. The number of moles of potassium ions will be 1.305.

Given that,

Now, calculating the number of moles in 343 mL,

Tripotassium phosphate is an ionic salt, which will dissociate into 3 in aqueous solution. It can be interpreted as 1 mole of tripotassium phosphate produces 3 moles of potassium ions.

Therefore, the 0.435 mole of will produce = 3 X 0.435 = 1.305 mole of K+ ions. The 343 mL of 1.27M solution consists of 1.305 moles of potassium ions.

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1 molar (M) solution is equivalent to the 1 mole of the compound present in 1 litter or 1000 mL of the solution. Thus 1.27 M (molar) solution represent 1.27 mole of K₃PO₄ present in 1000 mL of the solution. Thus 343 mL of the solution contains ×343 = 0.435 mole of K₃PO₄. Now K₃PO₄ is a strongly ionic salt which dissociate in solution as K₃PO₄=3K⁺ + . Thus 1 mole of K₃PO₄ produces 3 moles of K⁺ ions.

Thus, 0.435 mole K₃PO₄ produces (3×0.435) = 1.305 mole of K⁺ ion.

Thus 343 mL of 1.27 M K₃PO₄ solution contains 1.305 mole of K⁺ ions.      

answer is a

explanation:

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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