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3 years ago

what is the boiling point of the solution resulted from the dissolving of 32.5g of NaCl in 250.0g of water?

Chemistry

1 answer:

suter [353]3 years ago

8 0

Answer:

The boiling point of this solution is 102.28 °C

Explanation:

Step 1: 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

Step 2: 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

Step 3: 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

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