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Topic: Preparing 10% HCl  (Read 81716 times)

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Offline annietan

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Preparing 10% HCl
« on: April 29, 2012, 03:37:17 AM »
1) Prepare 100mL of 10%w/w HCl from 37%concentrated HCl
(density of 37%HCl =1.19g/ml)

2) Prepare 100mL of 10%w/v HCl from 37%
concentrated HCl(density of 37%HCl=1.19g/ml)

Is 10%w/v HCl the same as 10%w/w HCl?
Please show the calculation steps / formula for 1) & 2).

Offline Borek

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Re: Preparing 10% HCl
« Reply #1 on: April 29, 2012, 04:18:46 AM »
Is 10%w/v HCl the same as 10%w/w HCl?

No, these are different things.

Quote
Please show the calculation steps / formula for 1) & 2).

You have to show your attempts at solving the question to receive help. This is a forum policy.

You do know something about concentrations, don't you? You know the definition perhaps?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline annietan

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Re: Preparing 10% HCl
« Reply #2 on: May 01, 2012, 12:43:18 AM »
10% w/v means 10g of HCl in 100mL of solution and 10%w/w means 10g in 100g of solution.

1) For 10%w/v,
In 37% conc HCl,mass of HCl in 1000ml=1.19g/ml x1000ml x37%=440.3g.
mass of HCl in 100ml=44.03g.
10/44.03 x 100ml = 22.7ml of conc HCl. Add water to make up to 100ml to get 10%w/v HCl.

2) For 10%v/v,
Can you help me? What is the volume of 37% conc HCl (density 1.19g/ml) that I need to measure to make 100ml of 10%v/v HCl?


Offline annietan

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Re: Preparing 10% HCl
« Reply #3 on: May 01, 2012, 12:44:49 AM »
Sorry, typo error, can you help me with the volume of conc HCl required to make 100ml of 10%w/w HCl?

Offline Borek

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Re: Preparing 10% HCl
« Reply #4 on: May 01, 2012, 03:52:29 AM »
Sorry, typo error, can you help me with the volume of conc HCl required to make 100ml of 10%w/w HCl?

Hint: you will need density of the 10%w/w solution to calculate mass of 100 mL.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline crosemeyer

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Re: Preparing 10% HCl
« Reply #5 on: May 02, 2012, 02:13:41 PM »
Agree with Borek.  I don't think enough information is given to answer the question.

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