How much #NaNO_3# is needed to prepare 225mL of a 1.55 M solution of #NaNO_3#?

Answer 1

#"Approx. " 30* g#

#"Concentration"# #=#
#"Mass of stuff (g)"/"Molar mass of stuff (mol)"xx1/"Volume of solution (L)"#
The calculation gives me units of #mol*L^-1# as required.

Additionally, the item

#"Concentration"xx"Volume"# #=# #"Number of moles"#
#225xx10^-3*Lxx1.55*mol*L^-1# #=# #0.349*mol#
#0.349*molxx84.99*g*mol^-1# #=# #??g#.

Dimensionally, the calculation is consistent.

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Answer 2

To prepare 225 mL of a 1.55 M solution of NaNO3, you would need 35.0625 grams of NaNO3.

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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.

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.

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.

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.

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