A 150 mL of oxygen gas is collected over water at 20 °C and 758 torr. What volume will the same sample of oxygen occupy at STP when it is dry? The vapor pressure of water at 20 °C is 17.54 torr.

According to the ideal gas equalisation at a given temperature and pressure the capacity of the gas will be immediately proportionate to the volume of the gas.

It can be represented as:

P V = n RT

where, p = pressure, v = volume of the gas, n = amount, R = gas constant and T = temperature.

The volume of the gas would be 136.17 ml.

This can be estimated by:

At STP:

Now using the general gas equation:

V2 = 136.17 ml

Therefore, the volume of the gas sample when it got dried will be 136.17 ml.

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The volume would be; 136.17 ml

Explanation:

Volume V1 = 150 mL

Temperature T1 = 20°C + 273 = 293 K

Pressure  P1 = 758 - 17.54 = 740.46 torr

At STP;

Volume V2 = ?

Pressure P2 = 760 torr

Temperature T2 = 273 K

Using the general gas equation;

P1V1 / T1   = P2V2 / T2

Making V2 subject of formulae;

V2 = P1V1T2 / T1P2

Inserting the values we have;

V2 = 740.46 * 150 * 273  /  293 * 760

V2 = 136.17 ml

i believe the answer is 5,000