Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water, determine the constants m and b in the equation Molarity = mdensity + b that relates the NaCl molarity to the NaCl(aq) density. Take the density of water to be 1.00 g/mL and the molar mass of NaCl to be 58.5 g/mol.

I am having a very difficult time answering this question. There just doesn't seem to be enough information to determine the constants.

Question

Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water, determine the constants m and b in the equation Molarity = mdensity + b that relates the NaCl molarity to the NaCl(aq) density.  Take the density of water to be 1.00 g/mL and the molar mass of NaCl to be 58.5 g/mol.

I am having a very difficult time answering this question. There just doesn't seem to be enough information to determine the constants.

aaronq · Answer

http://openstudy.com/study#/updates/5246f87fe4b034054f3dd3cf

anonymous · Answer

Thank you for the response! I understand how you substituted the equation but I think I am missing something. When I plug in the values I get...

$$(58.5 gL/mol) \div (M _{NaCl}) =58.5gLmol \times x + Y$$

aaronq · Answer

$m_{NaCl}$ is mass, which is arbitrary, you can pick any mass you want. $M_{NaCl}$=molar mass of NaCl, which is 58.5 g/mol.

anonymous · Answer

Thank you that makes sense. I really appreciate you taking the time to answer my question.

aaronq · Answer

ok cool. no problem at all !

aaronq · Answer

btw, the way you would solve this, would be by, either, graphing, or finding the slope first, then the y intercept.

anonymous · Answer

Actually I may have spoken too soon. I used 10g Nacl for the arbitrary mass. Doing so i get the slope to be .17mol/L= 10g/L (x) + b. What would be the next logical step?

aaronq · Answer

you have to use 2 masses to get the slope, then use the slope and a point to get b

anonymous · Answer

Okay I got it for real this time. As part of the lab i had to compare the experimental value to the "simple model". I made my y axis concentration and my x-axis so I was getting an impossibly large slope in comparison to the slope for the simple model. Thanks for sticking with me!

aaronq · Answer

i'm glad you figured it out!

anonymous · Answer

Actually I may have spoken too soon. I used 10g Nacl for the arbitrary mass. Doing so i get the slope to be .17mol/L= 10g/L (x) + b. What would be the next logical step?

anonymous · Answer

Okay I got it for real this time. As part of the lab i had to compare the experimental value to the "simple model". I made my y axis concentration and my x-axis so I was getting an impossibly large slope in comparison to the slope for the simple model. Thanks for sticking with me!