We measured a certain amount of water using platinum weights. The weight of the water is $151,392 \mathrm{~g}$. What would have been the weight of the water if the measurement had been taken in a vacuum? The pressure of the air is $748 \mathrm{~mm}$, the temperature is $18^{\circ}$, the weight of $1 \mathrm{~cm}^{3}$ of dry air at $0^{\circ}$ and $760 \mathrm{~mm}$ is $0,001293 \mathrm{~g}$, the density of water at $18^{\circ}$ is 0,998654; the density of platinum is 21.
Let the weight of water in air be $q$, in a vacuum $q_{1}$, the normal atmospheric pressure $b$, the current atmospheric pressure $b_{1}$, the current temperature $t$, the normal density of air $s$, the current density $s_{1}$, the expansion coefficient of gases $a$, the density of water $S$, the density of platinum $S_{1}$, the volume of water $V$, and the volume of platinum $v$.

The difference in weight between the water and an equal volume of air on one pan of the balance, and the difference in weight between the platinum and an equal volume of air on the other pan, are in equilibrium, so

$$
V\left(S-s_{1}\right)=v\left(S_{1}-s_{1}\right)
$$

Since

$$
q_{1}=V S
$$

from the above equation,

$$
q_{1}=v \frac{S_{1}-s_{1}}{S-s_{1}} S
$$

On the other hand,

$$
v S_{1}=q
$$

and thus

$$
v=\frac{q}{S_{1}}
$$

and therefore

$$
q_{1}=q \frac{S_{1}-s_{1}}{S-s_{1}} \cdot \frac{S}{S_{1}}
$$

Finally,

$$
s_{1}=s \frac{b_{1}}{b(1+a t)}=0.001194
$$

and thus

$$
q_{1}=151.567 \text{ g}
$$

(Frigyes Riesz, Gymnasium VIII. grade, Győr.)

The problem was also solved by: Bernát Friedmann, Béla Grünhut, and István Szabó.