Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
Answer:
The molecular weight of monoprotic acid neutralized by 0.08133 M KOH is 202.105 grams.
Explanation:
KOH is considered as monoprotic base. So the neutralization reaction between monoprotic acid and KOH will under 1:1 molar ratio. Moles of monoprotic acid reacts with KOH will be equal to the moles of KOH.
Moles of KOH:
Molarity = [tex]\rm moles\;\times\;\frac{1000}{volume\; ml}[/tex]
0.08133 M = moles of KOH [tex]\rm \times\;\frac{1000}{16.4}[/tex]
moles of KOH = 0.00133 moles.
Equal moles of monoprotic acid is neutralized, i.e. :
0.00133 moles of monoprotic acid
moles = [tex]\rm \frac{weight}{molecular weight}[/tex]
0.00133 = [tex]\rm \frac{0.2688\;g}{molecular weight}[/tex]
Molecular weight of monoprotic acid = 202.105 grams.
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