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A #295mL# volume of solution that is #1.00molL^-1# with respect to #"ammonium chloride"# is diluted with a #2.25L# volume of fresh solvent. What is the resultant #[NH_4Cl]#?

Answer 1

Wyatt Atkins

Approx. #0.116*mol*L^-1#

Similar to all these kinds of issues,

#"concentration"="moles of solute"/"volume of solution"#

Therefore, we simply calculate volume and moles as needed:

#=# #(295*mLxx10^-3*L*mL^-1xx1.00*mol*L^-1)/(0.295*L+2.25*L)#

#=# #??mol*L^-1# with respect to #NH_4Cl#.

Given that the initial concentration is nearly tenfold diluted, the concentration makes sense.

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Answer 2

Cooper Adams

0.044 mol/L

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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.

A #295*mL# volume of solution that is #1.00*mol*L^-1# with respect to #"ammonium chloride"# is diluted with a #2.25*L# volume of fresh solvent. What is the resultant #[NH_4Cl]#?

A #295mL# volume of solution that is #1.00molL^-1# with respect to #"ammonium chloride"# is diluted with a #2.25L# volume of fresh solvent. What is the resultant #[NH_4Cl]#?