The benzene boiling temperature (c6h6) is 80.1ºc dissolving 36 g pentane, c5h12 at 500 g benzene increases the boiling point of the solution to 82.73ºc
a. consider the benzene boiling point constant. show calculations.
b. in dissolving 1.2 g of unknown solute in 50 g of benzene, a solution with a boiling point of 80.36ºc is obtained, which is the molar mass of the solute (assume that i = 1) (show calculations)

A)Boiling point constant of benzene = 2.63°C/m

B) 242.77 g/mol is the molar mass of the solute.

Explanation:

where,

=Elevation in boiling point

= boiling point constant od solvent= 3.63 °C/m

1 - van't Hoff factor

m = molality

A) Mas of solvent = 500 g = 0.500 kg

T = 80.1°C , =82.73°C

= 82.73°C - 80.1°C = 2.63°C

Boiling point constant of benzene = 2.63°C/m

B) Mass of solute = 1.2 g

Molar mas of solute = M

Mass of solvent = 50 g= 0.050 kg

i = 1

T = 80.1°C , =80.36°C

=80.36°C -  80.1°C = 0.26°C

M = 242.77 g/mol

242.77 g/mol is the molar mass of the solute.

they vary. it used to be meat, grain, bread, candy, beverages, donuts, and fruits/vegetables. now they added exercise in.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol