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

1 Answer
anor277
Jan 30, 2017

Approx. 100*mL

Explanation:

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 ADD THE CONCENTRATED ACID TO THE WATER NEVER THE REVERSE. WHY NOT?

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