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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? $\endgroup$ – Ivan Neretin Oct 11 '17 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 '17 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.Mc Oct 11 '17 at 20:33
  • $\begingroup$ @ Ivan, my question has evolved. The volume is indeed not the same, it is reduced. Which moves the SG to be greater than 1.009. Whereas the known SG is 1.0046, i.e. less than 1.009. which makes my reasoning even more wrong! $\endgroup$ – S.Mc Oct 21 '17 at 9:55
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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. $\endgroup$ – Ivan Neretin Oct 21 '17 at 10:49
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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

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

or

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

water.

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 $\pu{1.0046 g/cm^3}$.

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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.

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  • $\begingroup$ 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.Mc Oct 11 '17 at 20:39
  • $\begingroup$ @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 '17 at 12:14
  • $\begingroup$ Oh for heaven's sake.. In my original question, I stated that the saline was 9g of salt in a litre of water. That's a w/v statement right there, if you care to read it. $\endgroup$ – S.Mc Oct 20 '17 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.Mc Oct 20 '17 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.Mc Oct 21 '17 at 2:03
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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.

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  • $\begingroup$ 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.Mc Oct 21 '17 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.Mc Oct 21 '17 at 21:18
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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.

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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.

Known:

*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*

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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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