What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24