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How many grams of ammonium sulfate are needed to make a 0.25 M
solution at a concentration of 6 M?

Sagot :

Emilyluvsyaidn
198.12 g of ammonium sulfate are required to make 0.25 L of a solution with a concentration of 6M.

The amount in grams of ammonium sulfate are needed to make a 0.25L solution at a concentration of 6 M is 198 grams.

How do we calculate grams from moles?

Mass (W) in grams from moles (n) of any substance will be calculated by using the below equation as:

n = W/M, where

M = molar mass

And for this question moles will be calculated by using the below formula:

M = n/V, where

M = molarity of the solution = 6M

V = volume of the solution = 0.25 L

On putting values on the above equation, we get

n = (6)(0.25) = 1.5 moles

Now mass of 1.5 moles of ammonium sulfate will be calculated by using the first formula as:

W = (1.5mol)(132g/mol) = 198g

Hence required mass of ammonium sulfate is 198 grams.

To know more about mass & moles, visit the below link:
https://brainly.com/question/1358482

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