Answered: 0.33mol of Linux in 6.40 L of solution…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

0.33mol of Linux in 6.40 L of solution

Molarity= M

71.7g C2H6O in 2.30 L of solution

Molarity= M

12.57mg KI in 113.2ml of solutions

Molarity= M

Expert Solution

Step 1

Here we need to calculate the molarity of given solutions:

0.33mol of Linux in 6.40 L of solution
71.7g C2H6O in 2.30 L of solution
12.57mg KI in 113.2ml of solutions

molar mass of C₂H₆O = 46 g/mol

molar mass of KI = 166 g/mol

$m o l e s = \frac{m a s s (g)}{m o l a r m a s s (\frac{g}{m o l})}$

Step 2

Mathematically,

$M o l a r i t y (M) = \frac{m o l e s o f s o l u t e (m o l)}{v o l u m e o f s o l u t i o n (L)} = \frac{m a s s o f s o l u t e (g)}{m o l a r m a s s o f s o l u t e (\frac{g}{m o l}) \times V o l u m e o f s o l u t i o n (L)}$

1.) 0.33mol of Linux in 6.40 L of solution

moles of solute (Linux) = 0.33 mol

Volume of solution = 6.40 L

plugging values in the above formula, we get:

$M o l a r i t y (M) = \frac{0.33 m o l}{6.40 L} = 0.052 M$

Step by step

Solved in 4 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions