 # CHM 152 Solve simple questions Anonymous

### Question Description

hey there, plz solve each Question in file plz

### Unformatted Attachment Preview

Purchase answer to see full attachment

lecturernewt
School: University of Virginia  Attached.

Surname 1
Name:
Professor:
Course Information:
Due Date:
Concentration Calculations

Question 1
Beryllium Chloride molar mass= 79.9182 g/mol
A 0.050 M solution means there are 0.050 moles in 1-liter solution
So, if 0.050 moles = 1 L
?

= 125 mL (0.125 L)

0.125 𝐿 ×0.050 𝑚𝑜𝑙
1𝐿

= 6.25 × 10-3 mol

1 mole of beryllium chloride = 79.9182 g
6.25 × 10-3 mol =
=

?

0.00625 mol ×79.9182 g
1 mol

= 0.4995 g ≈ 0.5 g
Question 2
Molality- number of moles in a kg of solvent
Density of water =1000 kg/m3 = 1 kg/L = 1 g/mL,
125 mL = 0.125 L
Molality (m) = molarity (M) × kg

= 0.05 × 0.125 = 0.00625 moles of Beryllium Chloride

Molar mass of BeCl2 = 79.9182 g/mol
Therefore, mass of BeCl2 needed = 79.9182 × 0.00625 = 0.5 grams
Question 3
Volume of Solute

Concentration Percentage % (v/v) = Volume of Solution × 100

Surname 2
Volume of ethanol

Hence, 4.5 = 225 + Volume of ethanol × 100
Taking X to represent volume of ethanol (in millimeters),
4.5 (225 + X) = 100X
4.5 (225+X)
100

=X

0.045 (225 + X) = X
10.125 + 0.045X = X
10.125 = X – 0.045X
10.125 = 0.955X
X = 10.60 mL
Mass

Density = Volume
Mass = Density × Volume
Mass of ethanol (g) = 0.789 g/mL × 10 .60 mL = 8.37 g
Question 4
Making 1.25 molal NH4OH solution
NH4OH molecular mass = 35.04 g/mol
1.25 mol ×35.04 g NH4OH
1 mol NH4OH

= 43.81 g

Molal solution refers to 1 mole of a solute dissolved in 1000 g of a solvent.
To prepare 1.25 molal ammonium hydroxide, add 43.81 g solute (NH4OH) to a 1-liter volumetric
Question 5
NaNO3 molar mass = 84.9947g/mol
0.45

Number of moles of NaNO3 in 0.45 g = 84.9947 = 0.0053 mol
Now Molarity =

Number of moles
Volume

=

0.0053 mol
0.265 L

= 0.02 M

Question 6
Applying the dilution formula M1V1 = M2V2
M1 = 0.15 M

V1 = 125 mL

M2 = X

V2 =125 + 25 = 150mL

Surname 3
0.15 × 125 = 150X
X = 0.125 M ≈ 0.13 M
Molarity of the solution = 0.13 M
Question 7
M1 = 0.15 M

V1 = 100 mL

M2 = X

V2 = 150 mL

Making use of the dilution formula M1V1 = M2V2
0.15 × 100 = 150X
Thus X = 0.1 M
Molarity of the diluted solution = 0.1 M
Question 8
Dilution Formula M1V1 = M2V2
Variables are as follows:
M1 = 10 M

V1 = 250 mL

M2 = 0.05 M

V2 = X

10 × 250 = 0.05X
X = 50000 mL =50 liters
Question 9
M1V1 = M2V2
Variables are as follows:
M1 = 1.5 M

V1 = 345 mL

M2 = X

V2 = 250 mL

1.5 × 345 = 250X
X = 2.07 M ≈ 2.1 M
Question 10
M1V1 = M2V2
Variables are as follows:
M1 = 2.4 M

V1 = 500 mL

Surname 4
M2 = 1.0 M

V2 = X

2.4 × 500 = X
X = 1200 mL
1200ml is the combined volume of the solution. Since there is an initial 500 mL of water, the
amount that needs to be added will be given by:
1200 – 500 = 700 mL
Question 11
a. Molarity of NaOH in solution
Molar mass of NaOH = 39.997 g/mol
83

Number of moles of NaOH in 83 g = 39.997 = 2.075 moles
Density of water = 1 g/mL
Therefore, 750 mL of water =750 g of water
Molality =

1000 g × 2.075 moles
750 g

= 2.77 m

b. Percent by mass of NaOH in the solution
Total mass of solution = 750 + 83 = 883 g
83

% mass of NaOH = 883 ×100 = 9.96 %
Question 12
a. Molarity of methanol
Mas...

flag Report DMCA Review Anonymous
I was on a very tight deadline but thanks to Studypool I was able to deliver my assignment on time. Anonymous
The tutor was pretty knowledgeable, efficient and polite. Great service! Anonymous
I did not know how to approach this question, Studypool helped me a lot. Studypool 4.7 Trustpilot 4.5 Sitejabber 4.4 Brown University

1271 Tutors California Institute of Technology

2131 Tutors Carnegie Mellon University

982 Tutors Columbia University

1256 Tutors Dartmouth University

2113 Tutors Emory University

2279 Tutors Harvard University

599 Tutors Massachusetts Institute of Technology

2319 Tutors New York University

1645 Tutors Notre Dam University

1911 Tutors Oklahoma University

2122 Tutors Pennsylvania State University

932 Tutors Princeton University

1211 Tutors Stanford University

983 Tutors University of California

1282 Tutors Oxford University

123 Tutors Yale University

2325 Tutors