**Exercise 34: Determining Grams of Solute in Aqueous Solutions**Calculate the mass of solute in grams present in each of the following aqueous solutions:(a) 575 mL of a 2.00 M solution of nitric acid, HNO₃(b) 1.65 L of a 0.286 M solution of alanine, C₃H₇NO₂(c) 320 mL of a 0.0081 M solution of calcium sulfate, CaSO₄**Instructions:**Use the formula:\[ \text{mass of solute (g)} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)} \]No graphs or diagrams are present in this text.

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34 How many grams of solute are present in each of the following aqueous solutions? (a) 575 mL of a 2.00 M solution of nitric acid, HNO, (b) 1.65 L of a 0.286 M solution of alanine, C,H,NO, (c) 320 mL of a 0.0081 M solution of calcium sulfate, CaSO, 3.

34 How many grams of solute are present in each of the following aqueous solutions? (a) 575 mL of a 2.00 M solution of nitric acid, HNO, (b) 1.65 L of a 0.286 M solution of alanine, C,H,NO, (c) 320 mL of a 0.0081 M solution of calcium sulfate, CaSO, 3.

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...

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

**Exercise 34: Determining Grams of Solute in Aqueous Solutions**

Calculate the mass of solute in grams present in each of the following aqueous solutions:

(a) 575 mL of a 2.00 M solution of nitric acid, HNO₃

(b) 1.65 L of a 0.286 M solution of alanine, C₃H₇NO₂

(c) 320 mL of a 0.0081 M solution of calcium sulfate, CaSO₄

**Instructions:**
Use the formula:
\[ \text{mass of solute (g)} = \text{molarity (M)} \times \text{volume (L)} \times \text{molar mass (g/mol)} \]

No graphs or diagrams are present in this text.

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