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if 0.75 g of a monoprotic weak acid required 22.50 ml of 0.510 m naoh to titrate it, what is the molar mass of the acid? select one: 86 g/mol 51 g/mol 65 g/mol 33 g/mol 153 g/mol

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The molar mass of the acid: 65.36 g/mol

Calculate the molar mass of given acid?

Balanced chemical equation is:

HA + NaOH ---> NaA + H2O

let's calculate the mol of HA

volume , V = 22.5 mL

= 2.25×[tex]10^{-2}[/tex] L

using:

number of mol, n = Molarity×Volume

= 0.51×2.25×[tex]10^{-2}[/tex]

= 1.147×[tex]10^{-2}[/tex] mol

According to balanced equation:

mol of NaOH reacted = (1/1) × moles of HA

= (1/1)×1.147×[tex]10^{-2}[/tex]

= 1.147×[tex]10^{-2}[/tex] = number of moles of NaOH.

using:

number of mol = mass / molar mass

1.147×[tex]10^{-2}[/tex] mol = (0.75 g)/molar mass

molar mass = 65.36 g/mol

Therefore, The molar mass of the acid: 65.36 g/mol.

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