The boiling point elevation of an aqueous sucrose solution is foundto be 0.39 degrees celsius. What mass of sucrose (molar mass =342.30g/mol) is contained in 500.0g of the solution? Kb=0.512 degrees celsius/molality

A) 261 g sucrose
B)528 g sucrose
C)130 g sucrose
D)762g sucrose
E)223g sucrose

The mass of sucrose is 130 grams (option C)

Explanation:

Step 1: Data given

The boiling point elevation of an aqueous sucrose solution is foundto be 0.39 °C.

Molar mass of sucrose = 342.30 g/mol

Mass of the solution = 500.0 grams

Kb=0.512 °C/molal

Step 2: Calculate molality

Tb = i*Kb*m

⇒ Tb = The boiling point elevation of an aqueous sucrose solution = 0.39 °C.

⇒ i = The van't Hoff factor = 1

⇒ Kb = 0.512 °C/ molal

⇒ m = the molality = moles sucrose / mass water

0.39 °C = 1*0.512 °C/molal* molality

Molality =0.76 molal

Step 3: Calculate moles sucrose

Molality = moles sucrose / mass water

0.76 molal = moles sucrose / 0.500 kg

moles sucrose = 0.38 moles

Step 4: Calculate mass sucrose

Mass sucrose = moles sucrose * molar mass sucrose

Mass sucrose = 0.38 moles * 342.30 g/mol

Mass sucrose = 130 grams

The mass of sucrose is 130 grams (option C)

43.452

step-by-step explanation:

284 times .153 = 43.452

the dissolution of polymer can be thermodynamically described in 2 steps.