# How many grams of NaF are present in 1.3 L of a 0.78-M NaF solution?

### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

Chapter
Section

### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 12, Problem 12.5YT
Textbook Problem
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## How many grams of NaF are present in 1.3 L of a 0.78-M NaF solution?

Interpretation Introduction

Interpretation:

The grams of NaF present in given amount of 0.78MNaF solution are to be determined.

Concept Introduction:

Molarity is defined as the number of moles of solute per liter of solution. Mathematically, molarity (M) is calculated as:

Molarity(M)=Moles soluteLiters solution

The volume of solution should be in liters and the unit of molarity is M.

The relation between the numbers of moles (n) and molar mass is:

n=mmolar mass

Here, m is mass.

The molar mass of NaF is calculated as:

Molar mass of NaF= atomic mass of Na + atomic mass of F

### Explanation of Solution

Given information:

Volume of NaF is 1.3 L and molarity of NaF is 0.78 M.

The molarity is expressed as:

Molarity(M)=Moles soluteLiters solution

Rearrange the above expression for moles solute as:

Moles solute=M×Liters solution

Substitute 0.78 M for M and 1.3 L for Liters solution in the above expression as: Moles solute=0.78 M×1.3 L=1

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