# density calculation

by jimbo007
Tags: calculation, density
 P: 20 does anyone know how to calculate the density of a solution of 0.7M (say) NaOH solution. if so, could you please show me how its done i have looked everywhere on the net but to my extreme surprise i couldnt find it thanks jim
 P: 7 It's pretty much all about knowing how to convert units. 0.7 M = 0.7moles/L NaOH Mol Weight= 40g/mol So $$(0.7moles/L)*(40g/mol)=28g/L$$ and since density is usually in units of $$g/mL$$, $$(28g/L)(1L/1000mL)= .028g/mL$$
 PF Patron Sci Advisor P: 614 Hello Jim, Well, we may use the conventional formula d=m/v, and if your NaOH solution is 1L, then let me calculate the density: 1L, 0,7M NaOH=0,7 moles=0,7*40=28 grams of NaOH (in 1 L of solution) You may calculate the density for your needs; if you want it to be expressed as g/100 mL, then you may say that d=2,8 g/100 mL. If you want in as g/mL; then you may write 0,028 g/mL for the solution. If your solution is not 1 L, we can still use d=m/v. Just calculate the following: d {g/L}=(c {mol/L} * v {L}) * mw {g/mol} / v {L}; here, (c * v) * mw gives the mass of your NaOH solution (providing that v is given in liters), and further dividing it to v gives the density. So just multiply the concentration with molecular weight, that gives the density in g/L. Changing the units will lead to different expressions of density, you may even try these. Regards, chem_tr
P: 1

## density calculation

Neither of the above calculations is correct. They calculate "density" based only on the NaOH present but the solution is made of water and NaOH so the density must take into account both. There is no easy way to do this without a measurement. You must weight your solution and determine the amount of water present in addition to NaOH, as the exact amount of H2O needed to reach 1 liter cannot be calculated exactly, only approximated. For dilute solutions the density will be close to that of pure water,but for more concentrated solutions, there will be a significant difference. You should expect the density of be greater as more NaOH is dissolved in the solution.
P: 414
 Quote by ChemKen Neither of the above calculations is correct. They calculate "density" based only on the NaOH present but the solution is made of water and NaOH so the density must take into account both. There is no easy way to do this without a measurement. You must weight your solution and determine the amount of water present in addition to NaOH, as the exact amount of H2O needed to reach 1 liter cannot be calculated exactly, only approximated. For dilute solutions the density will be close to that of pure water,but for more concentrated solutions, there will be a significant difference. You should expect the density of be greater as more NaOH is dissolved in the solution.
At an even more pedantic level, there is no mention of this being a solution in water.
 Admin P: 21,722 At the most pedantic level, this thread is 7 years old and none of the original posters is still with us.

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