Having a massive brain fart SIMPLE molarity -> molality conversion

[soggybread]

Having a massive brain fart... SIMPLE "molarity -> molality" conversion...

This isn't even a homework question per se, I'm just trying to understand how the question converted the molarity to the molality value, via the density given. I just got a different value of molality when trying to solve for it manually... and am pretty sure I did something stupid.

Homework Statement

If the osmotic pressure of a 6.02×10^-2 M aqueous solution of Ca(NO₃)₂ was found to be 4.05 atm at 20°C. Note that the corresponding molality would be 6.10×10^-2 m, given that the density of the solution is 0.997 g/cm³...

Homework Equations

None.

The Attempt at a Solution

Okay... so the question tells me it has a molality of 6.10×10^-2 m,

So if I want to solve for that myself, I would start off with the given molarity, right? So it would be 6.02×10^-2 M then multiply it by the density of solution, 0.997g/cm³, or in other words, 0.997kg/1L ... and then we're done, right?

So:
6.02×10^-2 mol/L multiply 1L/0.997kg = 6.04x10^-2

... but the question tells me it's 6.10×10^-2 m ... and I have no idea why I'm not getting that value.

I don't know what's going wrong here... I'm sure I am missing something dead simple...

 

[Borek]

Molality <> molarity conversion isn't that easy, proper formula is much more complicated.

See example of concentration conversion - percentage to molarity. That's not exactly what you need, but the general approach outlined there will lead you to the molarity/molality formula. There is also a nice collection of conversion formulas on the same site.

 

[Blueshift5]

but I can't see it.
Hello,

It seems like you are on the right track with your calculations. However, I believe the discrepancy in your answer may be due to rounding errors. The molarity value of 6.02x10-2 mol/L may actually be slightly higher or lower due to the significant figures given in the problem. Additionally, the density value of 0.997 g/cm3 may also have some slight variation. These small differences in values can lead to a slightly different molality value when calculated manually.

In order to get the exact molality value of 6.10x10-2 m, it may be helpful to use more precise values for molarity and density. Also, make sure to keep track of significant figures throughout your calculations to avoid rounding errors.

I hope this helps and clarifies any confusion. Keep up the good work!