if 3.90 ml m l of vinegar needs 45.0 ml m l of 0.135 m m naoh n a o h to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.50 qt q t sample of this vinegar?

Respuesta :

180 grams of acetic acid are in a 1.50 qt q t sample of this vinegar.

No of moles of [tex]$\mathrm{NaOH}=45.0 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} \times \frac{0.110 \mathrm{~mol} \mathrm{NaOH}}{1 \mathrm{~L}}=0.00495 \mathrm{~mol} \mathrm{NaOH}$[/tex]At equival ence point, [tex]$n_{\mathrm{CH}_3 \mathrm{COOH}}=n_{\mathrm{NaOH}}=0.00495 \mathrm{~mol}$[/tex]

Concentration of [tex]$\mathrm{CH}_3 \mathrm{COOH}=\frac{0.00495 \mathrm{~mol}}{3.90 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}}=1.27 \mathrm{~mol} / \mathrm{L}=1.27 \mathrm{M}$[/tex]

Volume of [tex]$\mathrm{CH}_3 \mathrm{COOH}=1.50 \mathrm{qt} \times \frac{0.946353 \mathrm{~L}}{1 \mathrm{qt}}=1.42 \mathrm{~L}$[/tex]

Amount of [tex]$\mathrm{CH}_3 \mathrm{COOH}=1.42 \mathrm{~L} \times \frac{1.27 \mathrm{~mol} \mathrm{CH}_3 \mathrm{COOH}}{\mathrm{L}} \times \frac{60.05 \mathrm{~g} \mathrm{CH}_3 \mathrm{COOH}}{1 \mathrm{~mol} \mathrm{C \textrm {CH } _ { 3 } \mathrm { COOH }}}$[/tex][tex]$=108 \mathrm{~g} \mathrm{CH}_3 \mathrm{COOH}$[/tex]

Acetic acid, officially known as ethanoic acid, is an organic molecule with the chemical formula CH3COOH that is acidic and colourless. Apart from water and other trace ingredients, vinegar has at least 4% acetic acid by volume, making acetic acid its primary ingredient.

Acetic acid is a sour ingredient that can be used to sauce, vinegar, pickled vegetables, and as a raw material for spices. Acetic acid can be identified when employed as a food additive by its group name, substance name, or abbreviated term depending on the use.

To learn more about Acetic acid visit:https://brainly.com/question/15202177

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