A metal sphere weighing 15.45 g is added to 21.27 mL of water in a graduated cylinder. The water level rises to 24.78 mL. Calculate the density of the metal. An empty beaker weighs 32.4257 g. A 10-mL pipet sample of an unknown liquid is transferred to the beaker. The beaker weighs 40 1825 g when weighed with the liquid in it. Calculate the density of the unknown liquid.

Question

12-02-2020
Chemistry

contestada

A metal sphere weighing 15.45 g is added to 21.27 mL of water in a graduated cylinder. The water level rises to 24.78 mL. Calculate the density of the metal. An empty beaker weighs 32.4257 g. A 10-mL pipet sample of an unknown liquid is transferred to the beaker. The beaker weighs 40 1825 g when weighed with the liquid in it. Calculate the density of the unknown liquid.

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MrRoyal MrRoyal · Answer 1 · 2020-02-12T13:32:37+01:00

Answer:

1. Density of Metal = 4.40g/cm³

2. Density of Liquid = 0.77568g/cm³

Explanation:

1.

Given

Initial Volume of Water = Vi = 21.27mL

Volume of Water = VF = 24.78 mL

Mass of Sphere= 15.45 g

Density = Mass/Volume

Change in Volume = 24.78 - 21.27 = 3.51 mL

Density = 15.45/3.51

Density = 4.40g/ml

1ml = 1cm³

So, density = 4.40/cm³

2.

Initial Mass of Beaker = Mb = 32.4257g

Final Mass of Beaker = Mf = 40.1825 g

V = 10 mL

Density = D = ?

Change in Mass =: MF – MI= 40.1825 – 32.4257 = 7.7568 g

Density = Mass/Volume

Density = 7.7568g/10mL

Density = 0.77568g/mL

1mL = 1cm³

So, density = 0.77568g/cm³

sammieyemite sammieyemite · Answer 2 · 2020-02-12T13:46:26+01:00

Answer:

1. Density of metal sphere d = 4.40 g/mL

2. Density of the Liquid, d = 0.0776 g/mL

Explanation:

For the 1st question

Given: mass of metal sphere, m = 15.45 g, Initial volume of water, [tex]V_{o}[/tex] = 21.27 mL and final volume due to the added metal sphere, [tex]V_{f}[/tex] = 24.78 mL.

Volume of the metal sphere, V = [tex]V_{f}[/tex] - [tex]V_{o}[/tex]

V = (24.78 - 21.27) mL

V = 3.51

Since density, d = mass m ÷ volume V

Density of metal sphere d = 15.35 ÷ 3.51

d = 4.40 g/mL

For the second question:

Given: mass of empty beaker, [tex]m_{o}[/tex] = 32.4257 g; volume of the added liquid, V = 10 mL; mass of beaker added the liquid has been added, [tex]m_{f}[/tex] = 40.1825 g.

Therefore mass of the liquid m = [tex]m_{f}[/tex] - [tex]m_{o}[/tex]

m = (40.1825 - 32.4257) g

m = 7.7568 g

Density of the Liquid, d = mass m ÷ volume V

Therefore d = 7.7568 ÷ 10

d = 0.0776 g/mL