# Could You Please Help Me With Calculating Molarity And Molality?

## 5 Answers

Anonymous answered
2 moles of the ionic solute CaCl2 (molecular weight = 110) yields how many osmoles?
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Anonymous answered
The molarity of 245.0g of H2SO4 dissolved in 1.00 L of solution?
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Molarity=moles of the solute÷volume of solution in litres

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Anonymous answered
*Sorry for the previous post which looked like a jumbled mess. I really wish this thing supported html but this is the best I can do for now... Maybe this reply is a bit late now but I'll give it a shot anyways (its good practice for me anyway). These are the formulas you need to know:

Mass percent: 28.8% means there are 28.8g of FeCl3 in 100. G solution. The solvent is 100.g - 28.8g=71.2moles=mass/molar mass
Molality (m)=moles/kg of water*
*note: To find the mass of water, take mass of solution minus mass of solute. (The solute in this case is FeCL3)
Molarity (M)=moles/volume in liters
mole fraction=moles of solute/(moles of solute + moles of water)
Molar mass of FeCl3: 162.2g

Now to solve the problem:
First, you find the mass of the solution (not the same as the mass of FeCl3). Use the quantity 1.00L when the quantity of the solution is not given. In order to do that, convert 1.00L to grams of solution using the density given. (I used dimensional analysis though it looks weird when typed).
1.00L X (1000ml/1L) X (1.280g/1ml)= 1280.g solution
The easier way to think of this is simply changing the 1.280g into 1280.g Now take that number and multiply it by the mass ratio of FeCl3 to get the mass of FeCl3.
1280.g solution X (28.8 g FeCl3/100 g solution)=369 g FeCl3.
Find number of moles of FeCl3 by dividing mass FeCl3 by molar mass of FeCl3 (which when calculated is 162.2).
369 g FeCl3/162.2 g FeCl3= 2.27 mol FeCl3
Next, find the molarity (M) by moles/volume in liters. Remember that our volume is 1.00 L which we chose in the beginning.
2.27 mol FeCl3/1.00 L= 2.27 M FeCl3
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Anonymous answered
The molarity of a 0.75 L solution that contains 9.1 NaCl. NaCl has a molar mass of 58 g/mol
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