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Vedmedyk [2.9K]
2 years ago
10

Immunoglobulin G (IgG), formerly called gamma globulin, is a principal antibody in blood serum. A 0.470 g sample of immunoglobul

in G is dissolved in water to make 0.106 L of solution, and the osmotic pressure of the solution at 25 ∘C is found to be 0.733 mbar. Calculate the molecular mass of immunoglobulin G.
Chemistry
1 answer:
Nookie1986 [14]2 years ago
5 0

The molecular mass of the immunoglobulin G, given the data from the question is 1.53×10⁵ g/mole

<h3>How to determine the molarity</h3>

We'll begin by calculating the molarity of the immunoglobulin G. This is illustrated below:

  • Volume = 0.106 L
  • Temperature (T) = 25 °C = 25 + 273 = 298 K
  • Osmotic pressure (π) = 0.733 mbar = 0.733 × 0.000987 = 0.00072 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Van't Hoff factor (i) = 1
  • Molarity (M)

π = iMRT

M = π / iRT

M = 0.00072 / (1 × 0.0821 × 298)

M = 0.000029 M

<h3>How to determine the mole of immunoglobulin G</h3>
  • Molarity = 0.000029 M
  • Volume = 0.106 L
  • Mole =?

Mole = Molarity × volume

Mole = 0.000029 × 0.106

Mole = 3.074×10⁻⁶ mole

<h3>How to determine the molar mass of mmunoglobulin G</h3>
  • Mole = 3.074×10⁻⁶ mole
  • Mass = 0.470 g
  • Molar mass =?

Molar mass = mass / mole

Molar mass = 0.47 / 3.074×10⁻⁶

Molar mass = 1.53×10⁵ g/mole

Learn more about Osmotic pressure:

brainly.com/question/5925156

#SPJ1

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Answer:

D. 15.8atm

Explanation:

Given parameters:

Initial pressure = 13atm

Initial temperature  = 34°C = 34 + 273  = 307K

Final temperature  = 100°C = 100 + 273  = 373K

Unknown:

Final pressure  = ?

Solution:

To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.

The expression is shown mathematically below;

        \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

P and T pressure and temperature values

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 Insert the parameters and solve for T₂;

    \frac{13}{307}   = \frac{P_{2} }{373}  

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How many moles are there in 3.612 x 1024 molecules of CaO?
melamori03 [73]

The number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

<h3>Number of moles in the molecules</h3>

The number of moles in  3.612 x 10²⁴ molecules of CaO is calculated as follows;

6.02 x 10²³ molecules = 1 mole

3.612 x 10²⁴ molecules = ?

= (3.612 x 10²⁴ ) / (6.02 x 10²³ )

= 6 moles

Thus, the number of moles in  3.612 x 10²⁴ molecules of CaO is  6 moles.

Learn more about number of moles here: brainly.com/question/15356425

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On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
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Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
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The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

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