Medical Normal Saline (0.9% Saline) contains 9 g of salt dissolved in 1000ml of water.

Hence a litre of NS should weigh 1.009 kg.

Hence its specific gravity should be 1.009, but it’s only 1.0046. (I've measured it, and verified it on-line. It is apparently, a fact.) I thought I understood SG, it’s a simple concept but I am missing something.

EDIT: after more reading, I learned that dissolving NaCl in water results in a reduction in the original volume. So my question now becomes: We're dissolving NaCl in pure water at a temperature where 1.000kg mass of water has a volume of 1.000 litres. By my (erroneous) reasoning: Say the volume decreases by 1 ml, giving a final volume of 0.999 litres. So the density of the solution would be 1.009 kg/ 0.999 l = 1.010 kg/l Giving an SG of 1.010 (because water has a density of 1.000 kg/l, as above) If the volume didn't change, it would be 1.009, but it increases because of the reduction in volume. BUT: The known, published SG of 0.9% w/v NaCl solution is 1.0046. Can anyone explain why, please? or at least, why the SG is below 1.009, when it should be greater?

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    $\begingroup$ OK, you've got 9 g of salt and dissolved it in 1000ml of water. Who said the resulting volume is 1 litre? $\endgroup$ Oct 11, 2017 at 4:27
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    $\begingroup$ One thing to understand is that significant figures play a significant role in solving chemistry problems. A specific gravity of 1.0046 is only about 0.5% more that 1.000 which is pretty close to the density of pure water at body temperature. So if medical saline is 0.9% saline, is it really 0.90000000% or is it maybe really 0.904%? So in a lot of problems we'd assume that 9 grams of salt and 1000 ml of water give "about" 1000 ml of solution, but it isn't absolutely true. $\endgroup$
    – MaxW
    Oct 11, 2017 at 4:47
  • $\begingroup$ Thanks for the replies, but neither gives me an explanation. By my reasoning ( which is obviously faulty, and I'm trying to find out how), an SG of 1.0046 is HALF that of 1.009. That is, it would be 9 grams of salt dissolved in TWO litres of water. I know that, when the salt is dissolved, the volume will change infinitesimally, but it won't double!! The total weight WILL be 1.009 kg and surely, the volume will still be very close to 1000 ml, so the density will be 1.009 kg/l ( or g/ml) which when compared to that of water gives an SG of 1.009. Where is my reasoning incorrect? $\endgroup$
    – S.McNally
    Oct 11, 2017 at 20:33
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    $\begingroup$ Reduced relative to what? To the total volume of (1000 ml of water + 9 g of salt)? That't probably true, but the result is still greater than 1 litre. $\endgroup$ Oct 21, 2017 at 10:49
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    $\begingroup$ After reading all of the comments and answers, it must be EMPHASIZED that normal saline, or 0.9% (w/v) NaCl is 9 grams NaCl per 1000 ml OF SOLUTION, not 9 grams NaCl added to 1000 ml water. So, the final volume will be 1 liter, not more and not less. This was correctly pointed out by @aventurin. $\endgroup$
    – Dr. J.
    May 25, 2018 at 11:49

8 Answers 8


0.9% medical saline is defined to contain $\pu{0.90 g}$ sodium chloride per $\pu{1000 ml}$ solution at $\pu{22 °C}$. Its density is $\pu{1.0046 g/cm^3}$ at $\pu{22 °C}$.

The density of water is $\pu{0.9978 g/cm^3}$ at $\pu{22 °C}$.

If the volume would stay constant when you add $\pu{0.90 g}$ sodium chloride to $\pu{1000 ml}$ water, the resulting density would be

$$\pu{0.9978 g/cm^3} + \pu{0.0090 g/cm^3} = \pu{1.0068 g/cm^3}$$

The density of the 0.9% saline shows us that $\pu{1000 ml}$ of it contain

$$\pu{1004.6 g} - \pu{9.0 g} = \pu{995.6 g}$$


$$\frac{\pu{995.6 g}}{\pu{0.9978 g/cm^3}} = \pu{997.8 ml}$$


This means that the volume of the liquid phase increases when we add sodium chloride to water, what explains the difference in density from $\pu{1.0068 g/cm^3}$ to $\pu{1.0046 g/cm^3}$.


That’s not how percentage concentrations work. A $0.9~\%$ solution contains $\pu{9g}$ of salt per $\pu{991g}$ of water so that both add up to $\pu{1000g}$.

Also note that you cannot guess or estimate the density based on the density of water alone. It will change when stuff is dissolved.

  • $\begingroup$ Thanks, but NS is a w/v concentration, not w/w. It IS 9 g in 1000 ml. By my reasoning ( which is obviously faulty, and I'm trying to find out how), an SG of 1.0046 is HALF that of 1.009. That is, it would be 9 grams of salt dissolved in TWO litres of water. I know that, when the salt is dissolved, the volume will change infinitesimally, but it won't double!! The total weight WILL be 1.009 kg and surely, the volume will still be very close to 1000 ml, so the density will be 1.009 kg/l ( or g/ml) which when compared to that of water gives an SG of 1.009. Where is my reasoning incorrect? $\endgroup$
    – S.McNally
    Oct 11, 2017 at 20:39
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    $\begingroup$ @S.Mc If it’s w/v it should always be mentioned being one otherwise normal chemists would not know that they are dealing with something entirely non-standard. $\endgroup$
    – Jan
    Oct 12, 2017 at 12:14
  • $\begingroup$ In my original question, I stated that the saline was 9g of salt in a litre of water. That's a w/v statement. $\endgroup$
    – S.McNally
    Oct 20, 2017 at 20:52
  • $\begingroup$ All of which is irrelevant anyway. Why does a litre of solution of SG 1.0046 have 9 g of salt in it, and not 4.6g? $\endgroup$
    – S.McNally
    Oct 20, 2017 at 20:56
  • $\begingroup$ chemistry.stackexchange.com/users/7475/jan Jan, can you cast any light, I still can't understand why 1000ml of water with 9 g of salt, doesn't have an SG of 1.009 $\endgroup$
    – S.McNally
    Oct 21, 2017 at 2:03

If anybody is still interested: this is all about the partial/apparent specific volume of NaCl in water.

Your first and main mistake is assuming that the $\pu{9 g}$ of NaCl are dissolved in $\pu{1000ml}$ of water. Since the total volume is $\pu{1000ml}$ and the NaCl contributes some volume, it‘s less water (see below). Your second main mistake is assuming that adding $\pu{9g}$ of NaCl doesn’t change the volume of the solution, and thirdly, you confuse SG (relative density) with density.

How much volume the NaCl contributes is determined by its partial specific volume. At infinite dilution, you might expect the partial specific volume to be the reciprocal of the density, but it is often less, because the dissolved solute can take up some of the space between the solvent molecules, or it can be greater if the solute molecules tend to repel the solvent molecules, or the ions take up more space then they do when combined. The partial molar volume is greatly determined by intra-molecular forces. Hence it is for all intents and purposes incalculable and an empirical quantity.

Let‘s do the calculation:

You say the solution has an SG (relative density, i.e. relative to the density of water) of $1.0046$, but a cursory search leading to tables of density of NaCl solutions suggests this value is the actual density of the solution at 22 °C (I found a table of density which is cited as having been taken from "Perry's Chemical Engineers' Handbook" by Robert H. Perry, Don Green, Sixth Edition). If so, one liter has a mass of $\pu{1004.6g}$, and hence contains $\pu{1004.6 g} -\pu{9 g} = \pu{995.6g}$ of water.

The density of water at 22 °C is $\pu{0.99777 g ml-1}$, so this water has a volume of $\pu{995.6 g}/\pu{0.99777 g ml-1} = \pu{997.8ml}$.

Since the total volume is $\pu{1l}$, the $\pu{9g}$ of NaCl in the solution have resulted in a volume increase of $\pu{1000 ml} - \pu{997.8 ml} = \pu{2.2ml}$, hence NaCl at this dilution has an apparent specific volume of

$$\frac{\pu{2.2ml}}{\pu{9g}} = \pu{0.244 ml//g}$$

This agrees well with a simple article I found from a student („Solution Density and Partial Molar Volume as Functions of Concentration“ by Melissa Michaels) in which the author uses standard methods to determine the partial molar volume of NaCl at various concentrations based on experimental measurements of the density. NaCl in the solution with $\pu{10.34g}$ NaCl in $\pu{1000ml}$ of solution was shown to have a partial molar volume of about $\pu{13.68 ml mol-1}$ which would be a partial specific volume of $\pu{0.234 ml g-1}$.

The density of NaCl is $\pu{2.16 g ml-1}$ and the reciprocal of this is $\pu{0.4629 ml g-1}$, so the ions seem to be fitting in between the water molecules well and/or attracting them, which makes sense since NaCl is an ionic compound and water is a polar molecule.

These calculations and this statement are based upon the assumption that the true density of the solution at 22 °C is $\pu{1.0046 g ml-1}$.

  • 1
    $\begingroup$ Expression formatting reference: Basics/CH+Math expressions/Upright vs italic/Math tutorial/Advanced $\endgroup$
    – Poutnik
    Apr 16 at 12:59
  • $\begingroup$ To which aspect of the formatting are you referring? $\endgroup$
    – joshbuhl
    Apr 19 at 9:36
  • $\begingroup$ E.g. literal numerical values with units or mathematical/chemical expressions/equations. Take them as handy links. $\endgroup$
    – Poutnik
    Apr 19 at 9:59
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    $\begingroup$ E.g. if you write $\pu{1.123123E5 J K-1 mol-1}$ it is displayed as $\pu{1.123123E5 J K-1 mol-1}$. It also ensures the space between the literal value and the unit does not allow line breaks. $\endgroup$
    – Poutnik
    Apr 19 at 10:40

1000 mL of pure water weigh in at 997.047 g at 25 °C, ADD exactly 9 g to that and you have 1.006047 kg.

Remember density is temperature dependent and therefore 1000 ml of pure water may weigh more or less depending on which temperature the volume was measured at.

Unless you are working with super cooler pure water, you cannot reach a density of 1.000 g/mL.

  • $\begingroup$ Thank you Jeppe. I thought that might be the answer...but the change in density with temperature is the exact reason why Specific Gravity is used. Expressing the density of the solution, relative to the density of water at the same temperature (giving a dimensionless number) cancels out that factor. To simplify my calculations, I just took it to be at the temperature where water has a density of 1.000 g/ml so that the density of the solution is the same, numerically, as the SG. Nonetheless, to verify, let's say the solution is at 25C: $\endgroup$
    – S.McNally
    Oct 21, 2017 at 20:56
  • $\begingroup$ If the volume didn't change after adding salt, the density is 1006.047/1000 = 1.006047 g/ml and the SG is 1.006047 /0.997047 =1.009531 If the solution volume decreases by, say, 1 ml then the density would be 1006.047/999 = 1.00705405 g/ml and the SG would be 1.00705405/0.997047 =1.0100366 so the SG is greater than if the volume didn't change. The fact is that the SG is 1.0046. which is LESS than if the volume didn't change. Still doesn't explain it. $\endgroup$
    – S.McNally
    Oct 21, 2017 at 21:18
            Consider this statement from Wikipedia  .
                  “ A solution of  9 grams of sodium chloride (NaCl) dissolved in water, to a total volume of 1000 ml (weight per unit volume). The mass of 1 milliliter of normal saline is 1.0046 grams at 22 °C “(Wikipedia Saline (medical)}
     The 1.0046 g/ml is a density value not a specific gravity value.  Specific gravity is a relative density computation based upon a designated reference.   If the designated reference is  H2O at  4 oC , then the  specific gravity would be the same  1.0046 .  However, at  22 oC  the specific gravity of NS is  1.00684 using  0.99777  g/ml  as the designative reference. 

    A statistical error may be present – the value of 1.0046  appears to be a False  Null hypothesis that was taken to be True – the definition of a  Type II error .    The original source of 1.0046 is from a   1973   Handbook of Hydrology(1993 McGraw Hill ) book and then  implemented in an online calculator in 2010  ( ref #13 at Wikipedia Medical Saline )  .  In that reference  Normal  Saline density value of 1.0046 was calculated from fitting 4th degree polynomial for temperature and  salinity.  Analytical  methods for analyzing salt water densities in estuaries involving millions of cubic meters of water is considerably different than infusing one liter of  NS an 8-hour period ( less than 50 L of water) !     Also, polynomial fitting , especially high degree poly's , introduces error factors into the estimation , particularly for the values at the bottom a top of the range.   On average Sea water  is  3.5%. 
The Specific Gravity of Normal Saline is dependent upon the choice of the density value of NaCl. There seems to be some slight variation of density depending on the source;  2.16 ,  2.165  and  2.17 .   The density of NaCl would have to be  2.05 , far out of range , in order to generate a Sp Gr  of  1.0046 

     At  22  degree Celsius H2O Density 0.99777 

NaCl 2.16
((1000 - 9/2.16 )0.99777 + 9 )/(1000.99777) SpGr 1.00485

NaCl 2.165

((1000 - 9/2.165 )0.99777 + 9 )/(1000.99777) SpGr 1.00486

NaCl 2.17

((1000 - 9/2.17 )0.99777 + 9 )/(1000.99777) SpGr 1.00487

NaCl 2.05

((1000 - 9/2.05 )0.99777 + 9 )/(1000.99777) SpGr 1.0046

Based upon the discussion above. - this author opines that the specific gravity of NS ( @ 22 oC ) is hereby deemed to be 1.0049.


There is one error in the question. When sodium chloride is added to 1000 ml of water, the final volume is slightly less than the sum of the volume of sodium chloride plus 1000 ml but should not be less than 1000 ml.

Total volume = 4.17 ml (volume for 9g of sodium chloride based on density of sodium chloride = 2.16g/cm3) + 1000 ml = 1004.17ml

Total weight = 9 g sodium chloride + 1000 g water = 1009g

Density = 1009g/1004.17ml = 1.0048 g/L

Of course temperature has an effect on the water density.


The solution is a mixture of NaCl and Water. 0.9% Saline solution is the mass ratio between the NaCl and Water in the solution.


*Mass ratio (NaCl/Water) = 0.009

Density of NaCl = 2.165 g/mL

Density of Water = 1.0 g/mL

Total Volume of the Solution = 1000 mL*

Calculating the Properties of NaCl:

*Mass (NaCl) = 0.009x Mass (Water)

Volume (NaCl) = Mass (NaCl)/Density (NaCl)

Volume (NaCl) = (0.009x Mass (Water))/2.165*

Calculating the properties of Water:

Volume (Water) = Mass (Water)/Density (Water)

Calculate the Mass of Water: Total Volume = NaCl Volume + Water Volume

1000 = 0.004157 x Mass (Water) + Mass (Water)/Density (Water)

Mass (Water) = 995.86 g

Calculate the Mass of NaCl:

*Mass NaCl = 8.96 g

Total Mass (Solution) = 1004.82 g

Solution Density = 1.00482 g/mL

Solution SG = 1.00482*


Density of NaCl is indicated at 2.16g/cc. Given this 9.0g of NaCl added occupies 4.16667ml of the flask. This leaves 995.833333ml of water in the flask. Therefore:

995.833333ml water * 1g/ml = 995.833333g water + 9.0g NaCl in the flask = 1004.833333g in 1000ml = 1.00483g/ml


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