The volume of a sample of helium is 4.5 mL at 20.0°C and 203.0 kPa. What will its volume be in the figure?A.4.34 mLB.3.78 mLC.2.25 mLD.6.85 mL

The question seems to be incomplete as it does not specify the conditions (temperature and pressure) of the "figure" mentioned. However, if we assume that you want to know the volume at standard temperature and pressure (STP), we can use the ideal gas law to solve this problem.

The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin.

First, we need to convert the given temperature from Celsius to Kelvin. The conversion formula is K = °C + 273.15. So, 20.0°C = 293.15 K.

Next, we need to convert the given pressure from kPa to atm because the ideal gas constant R is usually defined with pressure in atm. The conversion factor is 1 atm = 101.325 kPa. So, 203.0 kPa = 203.0/101.325 = 2.003 atm.

Now we can plug these values into the ideal gas law to find the number of moles of helium. Rearranging the formula to solve for n gives us n = PV/RT. Substituting the given values gives us n = (2.003 atm)(4.5 mL)/(0.0821 L·atm/K·mol)(293.15 K) = 0.000329 mol.

Finally, we can use the ideal gas law again to find the volume of this amount of helium at STP. Standard temperature is 273.15 K and standard pressure is 1 atm. Substituting these values into the formula gives us V = nRT/P = (0.000329 mol)(0.0821 L·atm/K·mol)(273.15 K)/(1 atm) = 0.0074 L = 7.4 mL.

So, the volume of the helium sample at STP would be 7.4 mL. This is not one of the options given in your question, so there may be a mistake in the question or in the given options.