In this question about the amount of sulfuric acid in the atmosphere of Venus it is calculated that the amount of water in it's atmosphere is $9.6 \times 10^{15} \text{ kg }$ H$_2$O.
But the calculation didn't account for the difference in molar mass between water and CO$_2$, the main constituent of the atmosphere, so the amount above needs to be multiplied by a factor 18/44, giving a total amount of $3.93 \times 10^{15} \text{ kg }$ 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 would then become about $5.9 \times 10^{15} \text{ kg }$.
In one of the answers to that question it is calculated that there is a total of $9 \times 10^{13} \text{ kg }$ H$_2$SO$_4$ in the clouds of Venus.
Assuming that at least 75% by weight of the droplets is sulfuric acid, this means that no more than $2.25 \times 10^{13} \text{ kg }$ H$_2$O is in those clouds.
So almost all of the water must be outside the clouds !
The figure on page 14 of the dissertation of Yeon Joo Lee shows that beneath the cloud layers there is a thin haze region which has a thousandfold less mass loading than the clouds above it, so the amount of water there is negligible.
So almost all of the water must be above the clouds !
Is this true or must there be some miscalculations ?
Is there any publication that can confirm the question ?