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A #75mL# volume of #0.211molL^-1# #NaOH# was diluted to #125mL#. What is the new concentration?

Answer 1

Owen Ambrose

Approx. #0.127*mol*L^-1#..............with respect to #NaOH#

#"Concentration"="Moles of solute"/"Volume of solution"#

And the only equation we require for these kinds of issues is this one:

#i.e. "moles of solute"="concentration"xx"volume"#

And thus.

#"Concentration of NaOH"=(75xx10^-3*cancelLxx0.211*mol*cancel(L^-1))/(125xx10^-3*L)=??*mol*L^-1.#

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Answer 2

Naomi Aronson

The new concentration of the solution can be calculated using the formula for dilution:

[ C_1V_1 = C_2V_2 ]

Where: ( C_1 ) = initial concentration of the solution (0.211 mol/L) ( V_1 ) = initial volume of the solution (75 mL or 0.075 L) ( C_2 ) = final concentration of the diluted solution ( V_2 ) = final volume of the diluted solution (125 mL or 0.125 L)

[ (0.211 , \text{mol/L})(0.075 , \text{L}) = C_2(0.125 , \text{L}) ]

[ C_2 = \frac{(0.211 , \text{mol/L})(0.075 , \text{L})}{0.125 , \text{L}} = 0.1266 , \text{mol/L} ]

So, the new concentration of the solution is 0.1266 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 #75*mL# volume of #0.211*mol*L^-1# #NaOH# was diluted to #125*mL#. What is the new concentration?

A #75mL# volume of #0.211molL^-1# #NaOH# was diluted to #125mL#. What is the new concentration?