Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

What is the change in the freezing point of a solution made by dissolving 14.7 g of
C6H1206 into 150 ml of water? Answer in 2 decimal.
The density of water is 1.00 g/ml and and Kf = 1.86 °C/m

Sagot :

jbferrado

The change in freezing point is 1.02°C

The change in the freezing point (ΔTf) is calculated from the following equation:

ΔTf = i x Kf x m

Given:

i = 1 (Van't Hoff factor, is equal to 1 because the solute is a nonelectrolyte)

Kf = 1.86°C/m (Freezing point constant of the solvent)

m = molality

We have to calculate the molality (m), which is equal to the moles of solute in 1 kg of solvent:

m = moles solute/ kg solvent

The solute is C₆H₁₂O₆. Thus, we calculate the moles of solute by dividing the mass into its molecular weight (Mw):

Mw(C₆H₁₂O) = (12 g/mol C x 6) + ( 1 g/mol H x 12) + (16 g/mol O x 6) = 180 g/mol

moles solute = mass C₆H₁₂O/Mw(C₆H₁₂O) = 14.7 g/(180 g/mol) = 0.082 mol

The solvent is water. Its mass (in kg) is obtained from the volume and the density, as follows:

kg solvent = V x density = 150 mL x 1.00 g/mL x 1 kg/1000 g = 0.15 kg

Now, we calculate the molality (m):

m = moles of solute/kg solvent = 0.082 mol/(0.15 kg) = 0.546 m

Finally, we calculate the change in freezing point:

ΔTf = i x Kf x m = 1 x 1.86°C/m x 0.546 m = 1.02°C

To learn more about freezing point depression, you can visit this content:

https://brainly.com/question/4258941

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.