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lianna [129]
3 years ago
12

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

Chemistry
1 answer:
Tresset [83]3 years ago
4 0

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.

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Answer:

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Explanation:

To solve this problem we will first calculate the moles of Li₂O.                       One mole of given substance always contains 6.022 × 10²³ particles  which can be atoms, ions, molecules or formula units. This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                         Number of Moles  =  Number of Particles ÷ 6.022 × 10²³

Putting values,

                         Number of Moles  =  1.204 × 10²⁴ Particles ÷ 6.022 × 10²³

                         Number of Moles  =  2.0 Moles

Secondly, we will convert calculated moles to mass using following relationship.

                         Moles  =  Mass  /  M.Mass

Or,

                         Mass  =  Moles  ×  M.Mass

Putting values,

                         Mass  =  2.0 mol × 29.88 g/mol

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Answer:

Astatine is a halogen with several isotopes that all have short half-lives.

Which of the following combinations of mass number and neutrons are possible as isotopes of astatine? Choose one or more:

i) A = 211, n = 127

ii) A = 210, n = 125

iii) A = 220, n = 134

iv) A = 207, n = 122

v) A = 209, n = 124

Explanation:

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