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What the #"millimolarities"# with respect to sodium and chloride ions, of a #9.00g# mass of sodium chloride dissolved in a volume of #1L# of solution...?

Answer 1

Peyton Adams

#"Molarity"="Moles of solute"/"Volume of solution"#.

And here, we have a concentration of #154*mmol*L^-1#.

#"Molarity NaCl"=((9.00*g)/(58.44*g*mol^-1))/(1.000*L)=0.154*mol*L^-1#

But we were asked to supply #"millimolarity"# of #Na^(+)(aq)#.

Because sodium chloride reacts in solution to give equimolar #Na^+#, and #Cl^-# as the aquated ions....

#[Na^+]=[NaCl(aq)]=0.154*mol*L^-1#

Additionally.

#[Na^+]=0.154*mol*L^-1=0.154*cancel(mol)*L^-1xx10^3*mmol*cancel(mol^-1)=154*mmol*L^-1.#

And what is #[Cl^-]??#

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Answer 2

Axel Alvarado

The millimolarities of sodium and chloride ions in the solution would be both 154 millimolar (mM) when 9.00 g of sodium chloride is dissolved in 1 L of solution.

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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.

What the #"millimolarities"# with respect to sodium and chloride ions, of a #9.00*g# mass of sodium chloride dissolved in a volume of #1*L# of solution...?

What the #"millimolarities"# with respect to sodium and chloride ions, of a #9.00g# mass of sodium chloride dissolved in a volume of #1L# of solution...?