# CHM 152 Solve simple questions

*label*Science

*timer*Asked: Feb 19th, 2019

*account_balance_wallet*$20

### Question Description

hey there, plz solve each Question in file plz

## Tutor Answer

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

flask. Now add the solvent (water) up to the 1 L.

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...

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