Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

The freezing point of pure water is 0.0°C. How many grams of ethylene glycol (C2H6O2) must be mixed in 100.0 g of water to lower the freezing point of the solution to -4.3°C?

_______ g

Sagot :

pstnonsonjoku

The mass of the ethylene glycol used is 14.26 g

What is the freezing point?

The freezing point is the temperature at which a liquid is converted to a solid.

Given that;

ΔT = 0.0°C - ( -4.3°C) = 4.3°C

ΔT = 4.3°C

But we know that

ΔT = k m i

m = ΔT/k i

m = 4.3°C/1.86 °C/m * 1

m = 2.3 m

Now;

Let the mass of  ethylene glycol be m

Molar mass of  ethylene glycol = 62 g/mol

Mass of solvent = 100.0 g or 0.1 Kg

Then;

2.3= m/62/0.1

2.3 = m/62 * 0.1

m = 2.3 * 62 * 0.1

m = 14.26 g

Learn more about freezing point:https://brainly.com/question/3121416?

#SPJ1

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.