Answered

IDNLearn.com makes it easy to find precise answers to your specific questions. Join our Q&A platform to receive prompt and accurate responses from knowledgeable professionals in various fields.

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 :

Pstnonsonjokuidn

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

We are happy to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.