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 Message 1 of 2 in Discussion 
From: MSN NicknameN_2006  (Original Message)Sent: 4/14/2008 11:00 PM

The density of a 10% solution of sulfuric acid in water is 1.0687 g/ml. What is the molarity of this solution?
The answer is 1.09 M but I do not know how to get that value.
I calculated the molar mass of H2SO4 and then I assumed that the mass was 10 grams because it is 10percent.  is this right?

Second Question:
What is the minimum pressure required for a reverse osmosis water purification system at 25°C if the local water supply contains 0.300 g of CaCO3 per liter of solution?
The answer is supposed to be  0.147 atm, but I always get 0.0734 atm which is half.
I am using the PV=nRT equation.  P = MRT, calculating Molarity by taking 0.3 divided by molar mass of CaCO3 which is 100.06 g/mol, then I get
P = (0.003)(0.8206)(298)
But it is wrong.
Is it something to do because it is osmotic pressure?

Thanks in advance!



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Reply
 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 4/15/2008 5:17 AM
First Question
That's right, when you are given percentages, it's easiest to assume you have 100 g total, 10 g of H2SO4 and 90 g of water.  Molarity is moles of solute, H2SO4, per liter of solution.  The moles of H2SO4 is 10 g / MW of H2SO4 as you did.  For the liters of solution, first use the density formula V = m/d to convert 100 g of solution to mL, and then to L. 
 
Second Question
Don't forget that osmotic pressure is one of the colligative properties which mainly depends on the number of solute "particles" in the solution.  Since CaCO3 gives two particles, Ca2+ and CO32�?/SUP>, the molarity with respect to solute particles (the positive and negative ions) is 2 X M of CaCO3.  That will do it!
 
Steve