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kkurt [141]
3 years ago
10

A student has 100. mL of 0.400 M CuSO4(aq) and is asked to make 100. mL of 0.150 M CuSO4(aq) for a spectrophotometry experiment.

The following laboratory equipment is available for preparing the solution: centigram balance, weighing paper, funnel, 10 mL beaker, 150 mL beaker, 50 mL graduated cylinder, 100 mL volumetric flask, 50 mL buret, and distilled water.
a. Calculate the volume of 0.400 M CuSO4(aq) required for the preparation
b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO4(aq) from the 0.400 M CuSO4(aq)
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

<u></u>

  • <u>a. 37.5 mL</u>

<u></u>

  • <u>b. See the description below</u>

Explanation:

<u><em>a. Volume of 0.400 M CuSO₄(aq) required for the preparation</em></u>

In dissolutions, since the number of moles of solute is constant, the equation is:

          C_1\times V_1=C_2\times V_2

Substitute and solve for V₁

         0.400M\times V_1=0.150M\times 100ml

          V_1=0.150M\times 100.ml/0.400M=37.5ml

<u><em>b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO₄(aq) from the 0.400 M CuSO₄(aq)</em></u>

You will use the stock solution, the funnel, the buret, and distilled water.

i) Using the funnel, fill in the buret with with 50 ml of the stock solution, i.e. the 100. ml of 0.400 M CuSO₄(aq) solution.

ii) Pour 37.5 ml of the stock solution from the burete into the volumetric flask.

iii) Carefully add disitlled water to the 37.5ml of the stock solution in the volumetric flask until the mark (50 ml) on the volumetric flask.

iv) Put the stopper and rotate the volumetric flask to homegenize the solution.

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                   1.98 mole will contain  =  X Cl⁻ Ions

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Convert the composition of the following alloy from atom percent to weight percent (a) 44.9 at% of silver, (b) 46.3 at% of gold,
vesna_86 [32]

Answer:

Mass percentage of gold : 33.35%

Mass percentage of silver: 62.80%

Mass percentage of copper : 3.85%

Explanation:

N=n\times N_A

Where : N = Number of atoms

n = moles

N_A=6.022\times 10^{23} = Avogadro number

Let the total atoms present in the alloy be 100.

Atom percentage of silver = 44.9%

Atoms of of silver = 44.9% of 100 atoms = 44.9

Atomic weight of silver = 107.87 g/mol

Mass of 44.9 silver atoms = 107.87 g/mol ×  44.9 = 4,843.363 g/mol

Atom percentage of gold= 46.3%

Atoms of of gold = 46.3% of 100 atoms = 46.3

Atomic weight of gold = 196.97 g/mol

Mass of 44.9 gold atoms = 196.97 g/mol × 46.3 = 9,119.711 g/mol

Atom percentage of copper = 8.8 %

Atoms of of copper = 8.8% of 100 atoms = 8.8

Atomic weight of copper = 63.55 g/mol

Mass of 44.9 copper atoms = 63.55 g/mol ×  8.8 = 559.24 g/mol

Total mass of an alloy :4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol

Mass percentage of gold :

\frac{4,843.363 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=33.35 \%

Mass percentage of silver:

\frac{9,119.711 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=62.80\%

Mass percentage of copper :

\frac{559.24 g/mol}{4,843.363 g/mol + 9,119.711 g/mol +  559.24 g/mol}\times 100=3.85\%

7 0
3 years ago
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