A solution is prepared that is 23.44 wt% NaCl in water. What is the molality of the solution? MWNaCl = 58.443 g/mol)
Molality is the ratio between amount of substance of solute in moles and amount of solvent in kg.
Molality (M) = (moles of solute)/(mass of solvent)
Here the solute is NaCl and solvent is water.
23.44% wt means we have 23.44g of NaCl in 100ml of water.
For 1000ml of water we have `23.44xx10 = 234.4g` of NaCl.
Molic weight of NaCl = 58.443g/mol
moles of NaCl `= 234.4/58.443 = 4.01` moles
So we have 4.01 moles of NaCl in 1000ml of water.
Usually 1ml of water has a weight of 1g.
So in the solution we have 1000g = 1kg of water.
Molality = 4.01/1 = 4.01mol/kg
The molality of the NaCl solution is 4.01mol/kg
If the 23.44% is about the weight/volume ratio, then the answer can be found above.
If it is about the weight/weight ratio, then I suggest this answer:
23.44 % means the percentage of NaCl in the solution, which is composed of NaCl and water. So it is 23.44 % in 100 g of solution.
100 g=m (NaCl)+m (H2O)
m(H2O)= 100 g- m (NaCl)=100-23.44= 76.56 g
So, 1 kg of water would contain (1000/76.56)*23.44= 306 g NaCl
n(NaCL)= 306g/58.44g/mol= 5.24 moles
So the molality is 5.24 mol/kg