The normal boiling point of liquid methyl acetate is 331 k. assuming that its molar heat of vaporization is constant at 30.6 kj/mol, the boiling point of ch3cooch3 when the external pressure is 1.29 atm is

T_2=338.9026K

Boiling point of CH3COOCH3 at external pressure is 338.9026K

Explanation:

We are going to use Clausius-Clapeyron Equation:

Where:

P_2 is the external pressure

P_1 is the atmospheric Pressure=1 atm

ΔH is the heat of vaporization

T_2  boiling point of CH3COOCH3 at external pressure

T_1 normal boiling point of liquid methyl acetate

Now:

Boiling point of CH3COOCH3 at external pressure is 338.9026K

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm