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### Topic: 1 % HCL solution  (Read 80385 times)

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#### clinicalMT

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##### 1 % HCL solution
« on: August 18, 2004, 06:58:22 PM »
Thanks for the help on my molarity question... one more... could you show me how to make a 1 % HCL solution from 12 M HCL.

Clinical MT

#### justwonder

• Guest
##### Re:1 % HCL solution
« Reply #1 on: August 18, 2004, 11:08:26 PM »
Concentrated HCl has a molarity of approximately 11.6 (12M).

% solutions; note that liquid HCl is only 38% HCl, thus a 1% solution would require 2.6 mL of concentrated HCL (1/.38) per final volume of 100 mL.

38% xV1= 1% x 100ml
V1= 2,6ml.

use 2,6ml of 12M HCl solution and ad water to a final volum of 1ooml.
by the way here is a very good link you can use; http://www.rockystar.com/home/chemistry/chem_prep.htm

#### AWK

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##### Re:1 % HCL solution
« Reply #2 on: August 20, 2004, 03:06:28 AM »
Justwonder is true in this case but this is not a general method.
Let's start form beginning:
12 M HCl contains 12 moles = 12x36.461 g (437.53 g) in 1 L
1 % HCl shows density of 1.005 g/mL and in this case it can be rounded to density of water (usually 1.00 g/mL - but this is also approximation).

We need 10 g od pure HCl for 1L (1kg) of diluted solution.

So we need (10/437.61)x1000 mL of concentrated HCl (22.8 mL or 37.18 g) and water up to 1000 mL or 1005 g which is 967.8 g (or mL).

For 1L of 10 % HCl we need 228 mL 12M HCl (density 1.19 g/mL) weighted 371.8 g.
and enough water to 1000 mL or 1050 g (density of 10 % HCL is 1.05 g/mL) - 678,2 g (or mL) of water.
AWK

#### whywhywhy

• Guest
##### Re:1 % HCL solution
« Reply #3 on: September 17, 2004, 10:04:46 AM »

So we need (10/437.61)x1000 mL of concentrated HCl (22.8 mL or 37.18 g) and water up to 1000 mL .

hi AWK!

i tried to get your answer, but i can't. if 12% HCL has a density of  1.19 g/mL. then how come you get 37,18g 12%HCl? my calculation showed to be;

Mass= density x volume
mass=1.19 g/mL x 22,8ml
mass= 27,132 g
this mean we need 27, 132 g of 12%HCl solution.

hope that you can correct me so i can learn from my mistake.

thanks!