What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benzene by 6.28°C? (The boiling point elevation constant of benzene is 2.53°C•kg/mol.)

We need 17.1 grams of the solid

Explanation:

Step 1: data given

Molar mass of the solid = 46.0 g/mol

Mass of benzene : 150.0 grams

Boiling point elevation = 6.28 °C

(The boiling point elevation constant of benzene is 2.53°C•kg/mol.)

Step 2: Calculate molality of the solid

ΔT = i*Kb*m

⇒with ΔT = the boiling point elevation = 6.28 °C

⇒ with i = the van't Hoff factor of the solid = 1

⇒ with Kb = the boiling point elevation constant Kb of benzene : 2.53 °C /m

⇒with molality = TO BE DETERMINED

6.28 = 2.53 * m

m = 6.28 / 2.53

m = 2.48 molal

Step 3: Calculate number of moles of the solid

Molality = moles solide / mass benzene

2.48 molal = moles solid / 0.150 kg

moles solid = 2.48 molal * 0.150 kg

Moles solid = 0.372 moles

Step 4: Calculate mass of the solid

Mass solid = moles * molar mass

Mass solid = 0.372 moles * 46.0 g/mol

Mass solide = 17.1 grams

We need 17.1 grams of the solid

there are two significant figures.