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According to Wikipedia there are just 150 ppm of sulphur dioxide and 20 ppm of water in Venus' atmosphere. At the same time it is known that there is a considerable amount of sulphuric acid in the clouds. Venus' clouds are made of concentrated sulphuric acid, if I understand it correctly. This would suggest that the amount of sulphuric acid on Venus should be higher than the amount of water (otherwise sulphuric acid would absorb the water since it is strongly hygroscopic).

Also, if data cited on Wikipedia are from some spectrographic measurements, this may be affected by some bias: above the clouds there is probably a much lower concentration of sulphuric acid vapors than under the clouds.

I ask especially concerning possible colonization and local resource utilization on Venus. Two of the limiting substances are water and hydrogen, and a lot of $H_2SO_4$ would be helpful as a source of both hydrogen and water.

To be more specific with the question. 20 ppm of water means $9.6 \times 10^{15} \text{ kg } H_2O$ in total (Venus' atmosphere's mass is $4.8 \times 10^{20} \text{ kg}$). Is there less or more sulphuric acid?

Just for scale ... the amount of various sources of carbon on Earth follows:

 CO2 in Earth atmosphere     0.810e+15 kg
 Earth coal reserves         0.9e+15 kg (proven recoverable)
 Earth biosphere             1.9e+15 kg 
 CO2 in Earth sea water     36e+15kg
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First of all, thanks to @Robert Walker, who informed us about this dissertation of Yeon Joo Lee in his comment on the other answer.

In this article on page 14, we can see figure 1.3 with the vertical structure of Venus' cloud layers with mass loading profiles.

From this figure we can detect a cloud layer at about 48 to 50 km height with roughly a mass loading of 80 mg/m$^3$.

At 50 to 54 km height there is a mass loading of about 10 mg/m$^3$ and from 54-66 km i guess a mean mass loading of roughly 2 mg/m$^3$ would be suitable.

All these mass loadings from the 3 mentioned layers added up equals a roughly 2.8 km thick layer with a mass loading of about 80 mg/m$^3$.

To calculate the volume of this assumed layer we use Venus' surface area which is 4.6.10$^1$$^4$ m$^2$ and enlarge this area to 5.10$^1$$^4$ m$^2$ for the layer.

Accordingly the total weight of the droplets in this layer is:
5.10 $^1$$^4$ m$^2$ x 2.8.10$^3$ m x 8.10$^-$$^5$ kg/m$^3$ = 11.2.10 $^1$$^3$ kg.

Assuming 80% of H$_2$SO$_4$ in these droplets gives a total of about 9.10 $^1$$^3$ kg H$_2$SO$_4$.
According to the question the total amount of water is 9.6 x 10 $^1$$^5$ kg H$_2$O

So there is much less sulfuric acid than water !

Edit: According to the comment from @David Hemman below, the 20 ppm means a fraction of 20 moles of water to a million moles of mainly CO$_2$ gas, so the number in the question needs to be multiplied by 18/44, giving a total of $3.93 \times 10^{15}$ kg H$_2$O instead.

It should be noted that water vapor measurements by entry probes below the clouds have provided conflicting results, but that a consensus has emerged toward a mixing ratio of 30 +/- 10 ppm approximately constant in the whole 0-45 km range. (Composition of the atmosphere of Venus below the clouds, chapter 9)

Thus the total amount of water in the atmosphere of Venus would be: $5.9\times 10^{15}$ kg.

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  • $\begingroup$ Not a downvote, but close to it. This answer assumes that the 20 ppm value in the wikipedia article is correct and that it applies throughout Venus's atmosphere. The question also assumes that the 20 ppm is by mass rather than by volume. If the 20 ppm is by volume (which is typical), this alone more than halves the mass of the water in Venus's atmosphere. $\endgroup$
    – David Hammen
    Commented Sep 22, 2020 at 23:06
  • $\begingroup$ Micromole per mole is equivalent to ppmv (parts per million, by volume). It most definitely is not the same as parts per million, by mass. $\endgroup$
    – David Hammen
    Commented Sep 23, 2020 at 12:45
  • $\begingroup$ @DavidHammen One mole of water has the Avogadro number of molecules, and has a mass of about 18 grams. I don't see the connection with volume. $\endgroup$
    – Cornelis
    Commented Sep 23, 2020 at 13:25
  • $\begingroup$ @DavidHammen Ah, now I've got it. The molecules of water and CO2 are evenly mixed, so the ppm must be multiplied by the mass of water per mole and than divided by the mass of CO2 per mole. Thank you , I will change my answer. $\endgroup$
    – Cornelis
    Commented Sep 23, 2020 at 14:03
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OK, so I found something which more or less answers the question. Presentation Aerosols and Clouds on Earth and Venus from Owen Brian Toon of Department of Atmospheric and Oceanic Sciences and the abstract of BEYOND SULPHURIC ACID – WHAT ELSE IS IN THE CLOUDS OF VENUS?

In short the density of the clouds is about 1-50 $mg / m^3$ of liquid phase. Clouds are formed from 75-90% $H_2SO_4$. $6 \times 10^{13} \text{ kg }H_2SO_4$ is produced each year by photochemical processes.

But still it does not answer exactly what is the total amount.

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    $\begingroup$ Doesn't answer your original question, sorry, but just to say that in the clouds it's much denser. Assuming the droplets that form the cloud layer at 60 km are made of sulfiuric acid then it's an estimated 0.1 grams per cubic meter for the densest layer at 48.5 km, which compares favourably with fog levels on Earth of .05 g/m3 and is at the lower end of the density of cumulonimbus (thunderstorm clouds) at 0.1 - 0.3 g/m3. See page 14 of this paper: mps.mpg.de/phd/theses/… $\endgroup$
    – Robert Walker
    Commented Sep 25, 2016 at 12:28
  • $\begingroup$ @RobertWalker Thanks to you i could answer the question ! $\endgroup$
    – Cornelis
    Commented May 26, 2018 at 9:58

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