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How much 1.7M solution of #(NH_4)_2SO_4# do you have to use so that it contains 0.230 g of the substance?

Answer 1

Ellie Andrews

#"0.0011 L"#.

The mass of #1# mole of #(NH_4)_2SO_4# is:

#N xx 2 + Hxx4xx2 + S + Oxx4# #= 14.01 xx 2 + 1.008xx4xx2 + 32.07 + 16.00xx4# #= "132.14 g/mol"#

So, the number of moles of #(NH_4)_2SO_4# in #"0.230 g"# would be:

#"moles" = "mass"/"molar mass" = "0.230 g"/"132.14 g/mol" = "0.00187 mol"#

The question also tells us that the #(NH_4)_2SO_4# solution is #"1.7 M"#. This means that there are #1.7"# moles per liter.

#"1.7 moles"/"1 L"#

We can use this ratio to find the volume of solution which would have #"0.00187 mol"# of #(NH_4)_2SO_4#:

#"1.7 moles"/"1 L" = "1.7 × (0.00187/1.7) moles"/"1.0 × (0.00187/1.7) L"#

#= "0.00187 moles"/"0.0011 L"#

We'd need #"0.0011 L"# of the #(NH_4)_2SO_4# solution to get #"0.230 g"# of #(NH_4)_2SO_4#. :)

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Answer 2

Owen Ambrose

To prepare a solution containing 0.230 g of (NH4)2SO4, you would need to use 1.353 mL of a 1.7M solution of (NH4)2SO4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.