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iren [92.7K]
3 years ago
14

What mass of aluminum nitrate do you need to prepare 3.58L of a 1.77M Solution?​

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
8 0
You need a mass of 177
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There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Explanation:

Molarity of the solution = 2.20 M

Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}

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