Chemistry, 11.07.2019 06:30 QueenNerdy889

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

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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,

Molar solution of $\text {K}_{3} \text {PO}_{4}$ = 1.27 MVolume of solution = 343 mL

Now, calculating the number of moles in 343 mL,

$\dfrac{1.27}{1000}\times 343 &= 0.435 \; \text {moles of K}_{3}\text{PO}_{4}$

Tripotassium phosphate is an ionic salt, which will dissociate into 3 $\text{K}^{+} \text {and PO}_{4}^{3-}$ in aqueous solution. It can be interpreted as 1 mole of tripotassium phosphate produces 3 moles of potassium ions.

Therefore, the 0.435 mole $\text {moles of K}_{3}\text{PO}_{4}$ of will produce = 3 X 0.435 = 1.305 mole of K+ ions. The 343 mL of 1.27M $\text { K}_{3}\text{PO}_{4}$ solution consists of 1.305 moles of potassium ions.

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Answer from: sofiisabella10

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 $\frac{1.27}{1000}$ ×343 = 0.435 mole of K₃PO₄. Now K₃PO₄ is a strongly ionic salt which dissociate in solution as K₃PO₄=3K⁺ + $PO_{4}^{3-}$ . 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 from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

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