0.5008 g of an unknown triprotic acid, H3A, is dissolved in 47.3 mL of water and then titrated with 0.315 M NaOH. It takes 25.72 mL of the NaOH solution to completely neutralize the acid. What is the molar mass of this acid

Respuesta :

Answer:

The molar mass is "185.44 g/mol".

Explanation:

According to the question,

The moles of NaOH will be:

= [tex]\frac{0.315}{1000}\times 25.72[/tex]

= [tex]8.1018\times 10^{-3} \ moles[/tex]

Number of moles of an acid will be:

= [tex]\frac{8.1018\times 10^{-3}}{3}[/tex]

= [tex]2.7006\times 10^{-3} \ moles[/tex]

We know that,

⇒ [tex]Moles = \frac{Mass}{Molar \ mass}[/tex]

hence,

Molar mass of unknown acid will be:

= [tex]\frac{Mass \ in \ g}{Moles}[/tex]

= [tex]\frac{0.5008}{2.7006\times 10^{-3}}[/tex]

= [tex]185.44 \ g/mol[/tex]

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