Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?

Sagot :

JcAlmighty
The answer is 0.975 L

Volume = mol/Molarity

We have molarity (0.788 M) and we need mol and volume. Let's first calculate number of moles of CaCl2 in 85.3 g:

Molar mass of CaCl2 is sum of atomic masses of Ca and Cl:
Mr(CaCl2) = Ar(Ca) + 2Ar(Cl) = 40 + 2 * 35.45 = 40 + 70.9 = 110.9 g/mol

So, if 110.9 g are in 1 mol, 85.3 g will be in x mol:
110.9 g : 1 mole = 85.3 g : x
x = 85.3 g * 1 mole / 110.9
x = 0.769 moles

Now, calculate the volume:
V = 0.769/0.788
V = 0.975 L
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.