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### Topic: Solution & molality questions  (Read 4652 times)

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#### Aufbau89

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##### Solution & molality questions
« on: January 21, 2007, 08:36:26 PM »
Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84 grams per milliliter.  After dillution with water to 5.20-molar, the substance has a density of 1.38 grams per millilter and can be used as an electrolyte in lead storage batteries for automobiles.

a) Calculate the volume of concentrated acid required to prepare 1.00 L of 5.20-molar  H2SO4.
b) Determine the mass percent of  H2SO4 in the original concentrated solution.
c) Calculate the volume of 5.20-molar  H2SO4 that can be completely reacted with 10.5 grams of sodium bicarbonate, NaHCO3, in the reaction: 2NaHCO3 +  H2SO4 -->  2H20  + NasSO4 + 2 CO2
d) What is the molality of the 5.20-molar  H2SO4?

Work

a) (5.20M)(1.00L)= moles  H2SO4
= 5.20 moles  H2SO4

(5.20 moles)/(18.4M) = 0.28L  ?

-I'm sorry, this is a confusing chapter for me.  Can someone help me out with b,c, and d?  And was my A correct?  Sorry.  I really need help.
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#### Borek

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##### Re: Solution & molality questions
« Reply #1 on: January 22, 2007, 03:50:08 AM »
a. OK

b. Check http://www.chembuddy.com/?left=concentration&right=percentage-to-molarity - this is other way around, but you can easily solve final formula for C%.

c. Simple stoichiometry. Start calculating number of moles of substance in 10.5 g of sodium bicarbonate.

d. This one is a little bit more tricky. Look at the definition of molality - assuming you have 1L of solution, how can you calculate number of moles of acid and mass of solvent?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Aufbau89

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##### Re: Solution & molality questions
« Reply #2 on: January 28, 2007, 03:45:18 AM »
Ok thanks, I got it now.
Learn from yesterday, live for today, hope for tomorrow.
The important thing is to not stop questioning.
--Albert Einstein