what quantity of heat is required to vaporize 141 g of benzene at its boiling point 80.1 degrees celcius? heat of vaporization is 30.8 degrees celcius.
heat=massbenzene*Hv
YOur Hv is nonsense. Units can be Joules/kg, Joules/g, calories/gram, or such. degrees C isnot an acceptable unit for heat of vaporization.
my mistake. it's kj/mol
To find the quantity of heat required to vaporize benzene, you can use the equation:
q = m * ΔHvap
where:
q = heat required (in Joules)
m = mass (in grams)
ΔHvap = heat of vaporization (in J/g)
First, convert the boiling point of benzene from Celsius to Kelvin by adding 273.15:
T = 80.1°C + 273.15 = 353.25 K
Next, substitute the given values into the equation:
q = 141 g * 30.8 J/g
Calculate the quantity of heat:
q = 4342.8 J
Therefore, the quantity of heat required to vaporize 141 g of benzene at its boiling point is 4342.8 Joules.