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stepan [7]
2 years ago
9

The human eye has an osmotic pressure of 8. 00 atm at 37. 0 °c. what concentration (in moles/l) of a saline (nacl) solution will

provide an isotonic eyedrop solution?
Chemistry
1 answer:
Gelneren [198K]2 years ago
7 0

0.3147 concentration (in moles/l) of a saline (NaCl) solution will provide an isotonic eyedrop solution.

Isotonic eye drops

Because it might result in eye discomfort or tissue damage if it is not maintained, isotonicity is regarded as a crucial component of ophthalmic medicines. A few drops of blood are mixed with the test preparation before being examined and judged under a microscope at a magnification of 40. Isotonic solutions are those that have the same amount of water and other solutes in them as the cytoplasm of a cell. Since there is no net gain or loss of water, placing cells in an isotonic solution will not cause them to either shrink or swell.

We can calculate the osmotic pressure exerted by a solution using the following expression.

π = M . R . T

where,

π is the osmotic pressure

M is the molar concentration of the solution

R is the ideal gas constant

T is the absolute temperature

The absolute temperature is 37 + 273 = 310 K

π = M . R . T

8 = (X mol/L) . (0.082atm.L/mol.K) . 310 K = 0.3147 mol/L

To learn more about osmotic pressure refer:

brainly.com/question/5041899

#SPJ4

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<u>Solution</u> : Emperical Formula =  CH_{3}

Emperical Formula - A formula that gives the simplest whole number ratio of atom in a compound.

In the question percentages are given then assume that the total mass is 100grams so that the mass of each element is equal to the percentage given.

Given : 80.0g Carbon

            20.0g Hydrogen

First convert the given masses into moles.

80.0g\text{ C} \times \frac{1 g/mole \text{ C}}{12.0 g/mole \text{ C}}= 6.66\text{mole of C}

20.0g\text{ H} \times \frac{1 g/mole \text{ H}}{1 g/mole \text{ H}}= 20.0\text{mole of H}

Value of each mole divided by the smallest number of moles and then round off.

Moles ratio of the element are

\frac{6.66}{6.66}\text{mole of C}= 1

\frac{20.0}{6.66}\text{mole of C}= 3

This is the mole ratio of the elements and is repersented by the subscript in the emperical formula.

Emperical Formula =  CH_{3}

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The normal boiling point of liquid ethyl acetate is 350 K. Assuming that its molar heat of vaporization is constant at 34.4 kJ/m
Roman55 [17]

Answer:

337.22 K

Explanation:

Given that:

P₁ = 1 atm

T₁ = 350 K

P₂ = 0.639 atm

T₂ = ??? (unknown)

R(rate constant) =  8.34 J k⁻¹ mol⁻¹

Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.

Clausius-Clapeyron equation can be written as:

In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]

Substituting our values given; we have:

In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]

In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]

- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K

-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]

- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]

- 1.0825118*10^{-4} *350T_2 =T_2-350

- 0.037889T_2=T_2-350

350 = 0.03789T_2+T_2

350 =1.03789T_2

T_2= \frac{350}{1.03789}

T_2 = 337.22K

∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is <u>337.22</u> K.

5 0
3 years ago
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