Chemistry, asked by anchal01, 11 months ago

___ ml of HCl is required to prepare 750ml of 0.25M HCl

Answers

Answered by hrn21agmailcom

1

Answer:

6.84g

Explanation:

M = n/V

n = moles of HCl

n = Mass/Molar mass

n = Mass / 36.5

now

n = (Mass / 36.5) / 1L = 0.25

Mass = 0.25*36.5 = 9.125g

which means ....

1000mL contains 9.125g then

750ml contains 750*9.125/1000 = 6.84g

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