Relative Lowering of Vapour Pressure and Boiling Point Elevation

Which expression represents relative lowering of vapour pressure for a dilute solution?

• (p1° − p1)/p1° = n1/(n1 + n2)

• (p1° − p1)/p1° = n1/n2

• (p1° − p1)/p1° = n2/n1

• (p1° − p1)/p1° = n2/(n1 + n2)

For very dilute solutions (n2 << n1), which simplified relation holds?

• (p1° − p1)/p1° = n2/(n1 + n2)

• (p1° − p1)/p1° = n1/(n1 + n2)

• (p1° − p1)/p1° = n1/n2

• (p1° − p1)/p1° = n2/n1

Which formula links relative lowering of vapour pressure to masses and molar masses?

• (p1° − p1)/p1° = w1 M2/(M1 w2)

• (p1° − p1)/p1° = w2 M1/(M2 w1)

• (p1° − p1)/p1° = w1/(w2 M1)

• (p1° − p1)/p1° = M1/(M2 w1)

A non-volatile solute lowers vapour pressure because it

• raises atmospheric pressure

• adds volatile impurities

• increases solute activity

• reduces solvent mole fraction

Boiling occurs when vapour pressure of the liquid equals

• internal osmotic pressure

• Raoult’s Law constant

• external atmospheric pressure

• standard vapour pressure

Elevation of boiling point ΔTb is defined as

• Tb − Tb°

• p1° − p1

• Tb° − Tb

• Kb × w2/w1

For dilute solutions, elevation of boiling point is directly proportional to

• mole fraction of solvent

• density of solution

• vapour pressure of solute

• molality of solute

Which constant relates ΔTb to molality m?

• Kb (ebullioscopic constant)

• Ka (acidity constant)

• Ks (solubility constant)

• Kf (cryoscopic constant)

Given w2 grams of solute with molar mass M2 dissolved in w1 grams of solvent, molality m equals

• (w2 × 1000)/M1

• (w1 × M2)/w2

• (M2 × w1)/(1000 × w2)

• (1000 × w2)/(M2 × w1)

Using ΔTb = Kb m, the expression for molar mass M2 in terms of ΔTb, Kb, w1, w2 is

• M2 = (1000 × w2 × Kb)/(ΔTb × w1)

• M2 = (ΔTb × w1)/(1000 × w2 × Kb)

• M2 = (w1 × 1000)/(Kb × ΔTb)

• M2 = (Kb × ΔTb × w1)/1000