Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

<span> ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>

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How many mL will a 0.205 mole sample of He occupy at 3.00 atm and 200 K? Report your answer to the nearest mL.

Tcecarenko [31]

1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 3.00 atm

V= ?

n=0.205 mole

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$