The normal boiling point of methanol is 64.7â°c and the molar enthalpy of vaporization if 71.8 kj/mol. the value of î´s when 2.15 mol of ch3oh (l) vaporizes at 64.7â°c is j/k.

The value or amount of heat that is needed in order to vaporize the given amount of methanol (CH3OH) can be calculated through the equation,
    H = (n)(hf)

where H is the amount of heat
n is the given amount in mol, and 
hf is the molar heat of vaporization

Substituting the known values,

  H = (2.15 mol CH3OH)(71.8 kJ/mol)
  H = 154.37 kJ

ANSWER: 154.37 kJ

The molar enthalpy of vaporization is the amount of energy required to vaporize one mole of substance at a given temperature and pressure.

When we are given the amount of substance present, and the molar enthalpy of vaporization, we may simply use the formula:

ΔH = n * ΔH(vap) 

To find the enthalpy change occurring

ΔH = 2.15 * 71.8

The value of ΔH is 154.27 kJ.

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