295 mL of a 1.00 M ammonium chloride (#NH_4Cl#) solution is diluted with 2.25 L of water. What is the new concentration in molarity?

Answer 1

Ellie Andrews

Approx. #0.13*mol*L^-1#

#"Concentration"# #=# #"Number of moles"/"Volume of solution"#.

Initially, we have #0.295*Lxx1.00*mol*L^-1# #NH_4Cl# #=# #??*mol#

Finally this molar quantity is dissolved in a #2.25*L# volume.

#"Concentration"# #=# #(0.295*Lxx1.00*mol*L^-1)/(2.25*L)# #=# #??mol*L^-1#

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Answer 2

Peyton Adams

Apply the dilution formula: (M_1V_1 = M_2V_2). Substitute values: (1.00 M \times 295 mL = M_2 \times (295 mL + 2250 mL)). Calculate the new concentration in molarity.

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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.