# How many grams of NaOH are needed to prepare 0.500m solution using 800.o grams water and other problems

Submitted by ^^, on Sat, 01/10/2009 - 02:45

i'm having troubles with my homework.. can you help me?  :)

1. How many grams of NaOH are needed to prepare 0.500m solution using 800.o grams water?

2. What is the molality of a 0.300 M acetic acid(C2H4O2) solution with a density of 1.004g/mL?

3. What is the molality of a 2.75 mol Kl dissolved in 2.00L water?

pls help me...  ??? tnx.. ^^,

1) m= molality = moles solute/kg solvent

You are asked to calculate grams of NaOH. You're given the molality of the solution and the mass of the solvent (water). First thing's first, convert grams of water to kg of water
800g H20= 0.8kg H20

Now you're ready to go. Plug the values you have into the formula for molality (given above).
0.5m = x/0.8kg

Solve for x, which is the moles of NaOH in the solution. From there its just basic stoicheometry-convert moles to grams and you're done

2) Since you're not given a specified volume of solution, you can make up your own number for the volume. For convenience, we'll use 100ml of solution. Converting this value to L, we get 0.100L. You now must find the moles of acetic acid in the solution:
0.300M= moles of acetic acid/0.100L

moles of acetic acid= 0.03moles = 1.8g

the mass of the solution is 100ml X 1.004g/ml = 100.4g
And the mass of the solvent (water) is therefore,
mass of solution- mass of solute = mass of solvent
100.4g -1.8g = 98.6g H2O
Convert that mass to kg, which is 0.0986 kg

finally, calculate the molality: 0.03 mol acetic acid/ 0.0986kg = molality

3) This problem is a piece of cake. All you need to know is that the density of water is about 1.0g/ml. Since you have 2.0L, or 2000 ml, you have 2000g of H2O. This is converted to 2.0kg of H2O. You're already given moles of KI, so just plug them into the molaity formula and you're good to go!