# American Commercial College of Texas Chemistry Quiz

Anonymous

### Question Description

Hello, I have 8 questions! need to solve them as shown in the sample, i'll upload all the pictures when someone accepts the question, cuz I can only upload 5 atm.

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achiaovintel
School: Carnegie Mellon University

Attached.

1

Chemistry Questions
Name
Institution

CHEMISTRY QUESTIONS

2

Question 32
A solution of ethanol C2 H5 OH in water is prepared by dissolving 75 mL of ethanol
(density=0.79g/cm3) in enough water to make 250 mL of solution. What is the molarity of the
ethanol solution?
Solution
mass of ethanol used = ρ × v = 0.79
molar mass of ethanol = 46.08

g
× 75 cm3 = 59.25 g C2 H5 OH
cm3

g
C H OH
mol 2 5

Number of ethanol moles in 250 mL solution
Nummber of moles = 59.25g C2 H5 OH ×

1 mol
= 1.2858 moles C2 H5 OH
46.08 g

Concentration
Molarity = 1.2858 moles C2 H5 OH ×

1000mLsoln
1
5.1432 moles C2 H5 OH
×
=
L soln
250 mLsoln
L soln

𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓. 𝟏𝟒𝟑𝟐 𝐌
Question 34
Calculate the concentration of all ions present in each of the following solutions of strong
electrolyte.
b) 0.3 mole of barium nitrate in 600.0 mL of solution
Solution

Molarity Ba(NO3 )2 = 0.3 molBa(NO3 )2 ×
=

1
1000mL soln
×
600 mLsoln
L soln

0.5 mole Ba(NO3 )2
= 0.5 M
L soln

CHEMISTRY QUESTIONS

3

Ba(NO3 )2 → Ba2+ + 2NO3−
𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐁𝐚𝟐+ = 𝟎. 𝟓 𝐌
𝐂𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐍𝐎−
𝟑 = 𝟐 × 𝟎. 𝟓 𝐌 = 𝟏. 𝟎 𝐌
Question 38
If 10g of AgNO3 is available, what volume of 0.25 M AgNO3 solution can be prepared?
Solution
AgNO3 molar mass = 169.87

AgNO3 moles = 10 g ×

Volume of AgNO3 =

g
mol

1 mol
= 0.05886 mole AgNO3
169.87 g

L soln
1000 mL soln
×
× 0.05886 mol AgNO3
0.25 mol AgNO3
L soln

Volume of AgNO3 = 235.47 mL soln
𝐀𝐧𝐬𝐰𝐞𝐫 = 𝟐𝟑𝟓. 𝟒𝟕 𝐦𝐋
Question 42
How would you prepare 1.00 L of 0.5M solution of each of the following?
a) H2SO4 from concentrated (18 M sulfuric acid)
Solution
Moles of H2SO4 to be prepared
=

0.5 mole H2 SO4
× L soln = 0.5 moles H2 ...

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