# If 40.0mL sample of 0.130 M NaCl is diluted to 100.0 mL, hat is the molality of the resulting solution

I have this question about molality, and I am confused about which numbers to use.

A 40.0mL sample of 0.130 M NaCl is diluted to 100.0 mL. What is the molality of the resulting solution?

molality = moles of solute/mass of solvent(kg)

molality = 0.130M/1000(kg) x this is where i am confused on which numbers to use?

### In theory, it does not seem

KMST Sun, 09/04/2011 - 14:11

In theory, it does not seem solvable with the information, supplied. In practice, here is how I would solve it.

0.130 M = 0.013 moles solute/liter solution

Presumably a mass of NaCl of well known purity was accurately measured, dissolved and diluted to a volume accurately measured at 20.0°C in class A volumetric glassware to get such a solution. Volumetric class A glassware would have tolerances that keep your errors at 0.1% or less (not much less). That is usually all the precision we need. I do not know who would check the temperature of a solution, but it would take 5°C to change the density of water by 0.1%, and we assume the lab is at about 20°C, and that we would sense 5°C differences in temperature.

On diluting 40.0 mL of 0.130M solution to 100.0 mL, the final concentration would be 0.0520 moles solute/L solution.

That is 3.04g/L (0.0520 X 58.45 = 3.04), or 0.3%.

In the absence of any other specific information, we assume that the solvent is water.

Since the solution is very dilute, we could assume that the density is 1.000 g/mL = 1.000 kg/L

(My handbook says a 1% NaCl solution has a density of 1.004 g/mL at 25°C and 1.007 g/mL at 10°C

So 1.000 L of the final solution, would have a mass of about 1.000 kg (maybe a tiny bit more), including 0.003 kg (3 g) of NaCl, with the rest being solvent (water). There would be 0.0520 moles in that 1.000 L of solution. What is the mass of water in that 1.000 L of solution? Probably a bit more than 0.997 kg. Probably less than 1.004 kg. I would say to take 1.000 kg as a good estimate, and say you have 0.0520 moles of NaCl in 1.000 kg of water, and your moolality is 0.0520m. Your mass for a 1.000L of that solution would be 1.003 kg