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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 73p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 73p

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# Solved: A chemist wants to make 2.5 L of a 0.100 M KCl solution. How much KCl in grams ISBN: 9780321910295 34

## Solution for problem 73P Chapter 13

Introductory Chemistry | 5th Edition

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Problem 73P

Problem 73P

A chemist wants to make 2.5 L of a 0.100 M KCl solution. How much KCl in grams should the chemist use?

Step-by-Step Solution:
Step 1 of 3

Solution: here, we are going to calculate the mass of KCl required to make the solution.

Step1:

Molarity(M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

Moles of solute

Molarity = -------------------------------         ------(1)

Volume of solution in litre

Step2:

In the above problem,

Volume of the KCl solution = 2.5 L

Molarity of the KCl solution = 0.100 M

Substituting the values in equation (1), we get,

0.100 M = moles of KCl / 2.5 L

Therefore, moles of KCl = 0.100 M x 2.5 L

= 0.25 mol

Step3:

1 mol KCl = molar mass of KCl

0.25 mol KCl = 0.25 x molar mass of KCl

= 0.25 x 74.55 g/mol

= 18.64 g.

Thus, the required mass of KCl is 18.64 g.

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Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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