Answer:
The boiling point of this solution is 102.28 °C
Explanation:
<u>Step 1:</u> Data given
Mass of Nacl = 32.5 grams
Molar mass of NaCl = 58.45 g/mol
Mass of Water = 250 grams
Boiling point of water = 100°C
<u>Step 2: </u>Calculate number of moles
Number of moles = mass of NaCl / Molar mass of NaCl
Number of moles = 32.5 grams /58.45 g/mol = 0.556 moles
<u>Step 3:</u> Calculate molality
Molality = Number of moles / mass of water
Molality = 0.556 moles / 0.250 kg of water
Molality = 2.224 molal
NaCl releases twice as many moles of ions, the total molality of this solution is also twice: 4.448 molal
Step 4: Calculate boiling point
dT =( 0.512 C / molal)*4.448 molal)
dT = 2.28
The boiling point of this solution is 100 °C + 2.28 °C = 102.28 °C
