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How many mL of 10.0 M HCI are needed to prepare 500. mL of 2.00 M HCl?

Answer 1

Wyatt Atkins

Approx. #100*mL#

The product, #"concentration"xx"volume"=mol*L^-1xxL#, clearly has units of #mol#, i.e. #mol*cancel(L^-1)xxcancelL=mol#.

And thus #C_1V_1=C_2V_2#, and given the equality we can use units of #mL# for volume.

#V_1=(C_2V_2)/C_1# #=# #(500*mLxx2.00*mol*L^-1)/(10.0*mol*L^-1)#

#=??mL#

And you never add the reverse to the water—why not add the concentrated acid?

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Answer 2

Mateo Aguilar

To prepare 500 mL of 2.00 M HCl, you would need to dilute the 10.0 M HCl solution. You can use the formula for dilution: C1V1 = C2V2.

Given: C1 = 10.0 M (initial concentration) V1 = unknown (volume of 10.0 M HCl to be used) C2 = 2.00 M (desired concentration) V2 = 500 mL (final volume)

Using the formula: (10.0 M)(V1) = (2.00 M)(500 mL)

Solving for V1: V1 = (2.00 M)(500 mL) / 10.0 M V1 = 100 mL

So, you would need 100 mL of 10.0 M HCl to prepare 500 mL of 2.00 M HCl.

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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.