Titration of 0.4089 g of an unknown monoprotic acid dissolved in 25.00 ml of wat

Question

Titration of 0.4089 g of an unknown monoprotic acid dissolved in 25.00 ml of water requires 28.45 ml of 0.1521 M NaOH to reach the endpoint. What is the molar mass of the acid?
A) 107.5 g/mol B) 2.718 g/mol C) 94.49 g/mol D) 2.186 g/mol E) 0.01058 g/mol Titration of 0.4089 g of an unknown monoprotic acid dissolved in 25.00 ml of water requires 28.45 ml of 0.1521 M NaOH to reach the endpoint. What is the molar mass of the acid?
A) 107.5 g/mol B) 2.718 g/mol C) 94.49 g/mol D) 2.186 g/mol E) 0.01058 g/mol
A) 107.5 g/mol B) 2.718 g/mol C) 94.49 g/mol D) 2.186 g/mol E) 0.01058 g/mol

Explanation / Answer

ans ) B

solution : M1V1 = M2V2

(NaOH) 0.1521*28.45 = ?*25 (monoprotic acid)

concentration of monoprotic acid = 0.173 M

molar mass of acid

M = wt/mwt *1000/25

0.173 = 0.4089/? *1000/25

mwt = 94.49 g/mol

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