We appreciate your visit to What is the freezing point of 16 1 g of NaCl in 139 mL of water. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Final answer:
The freezing point of a 16.1 g NaCl in 139 mL water solution would be -2.46 °C.
Explanation:
The freezing point of a solution is lowered when a solute, such as NaCl, is added to a solvent, such as water. To calculate the freezing point depression, we can use the formula ΔT = Kf × m, where ΔT is the freezing point depression, Kf is the cryoscopic constant, and m is the molality of the solution.
First, we need to calculate the molality of the solution. Molality (m) is the moles of solute divided by the mass of the solvent in kilograms. The molar mass of NaCl is 58.44 g/mol.
m = (16.1 g NaCl / 58.44 g/mol) / (0.139 kg H2O) = 1.32 mol/kg
Next, we can use the cryoscopic constant (Kf) for water, which is 1.86 °C/m, to calculate the freezing point depression:
ΔT = (1.86 °C/m) × (1.32 mol/kg) = 2.46 °C
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution would be 0 °C - 2.46 °C = -2.46 °C.
Learn more about Freezing point depression here:
https://brainly.com/question/30093051
Thanks for taking the time to read What is the freezing point of 16 1 g of NaCl in 139 mL of water. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
