The Î""G°′ of the reaction is −7.180 kJ·mol−1. Calculate the equilibrium constant for the reaction at 25 °C.
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Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles =
= = 0.001245 moles
Concentration of KHP (C1) in litres = n/v
= = 0.062 mol/L
We know that:
=
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>