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lora16 [44]
3 years ago
8

A student is conducting molecular weight determination of an unknown diprotic acid. He first prepared a 250 mL of 0.15 M NaOH (3

9.997 g/mol) titrant solution, which he then standardized with 0.7625 g of KHP (204.22 g/mol) which consumed 26.32 mL of the titrant. After standardization, he weighed 0.0996 g of unknown diprotic acid and the titration reached the endpoint at a volume of 11.68 mL. What is the standardized concentration of titrant?
Chemistry
1 answer:
sweet [91]3 years ago
7 0

Answer:

0.0710

Explanation:

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Answer:

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Explanation:

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n =  \frac{N}{L} \\

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L is the Avogadro's constant which is

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From the question we have

n =  \frac{3.01 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{3.01}{6.02}  = 0.5 \\

We have the final answer as

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See answer below

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