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IrinaK [193]
2 years ago
12

Naproxen is a commercially important anti-inflammatory agent that can be isolated from the thyroid gland. A solution of 1.138 g

of naproxen in 25.00 g benzene has an osmotic pressure of 4.00 atm at 20°C. The density of benzene is 0.8787 g/mL at this temperature. Calculate the molar mass of naproxen, assuming it remains intact upon dissolution and the density of the solution equals the density of pure benzene. Naproxen is a commercially important anti-inflammatory agent that can be isolated from the thyroid gland. A solution of 1.138 g of naproxen in 25.00 g benzene has an osmotic pressure of 4.00 atm at 20°C. The density of benzene is 0.8787 g/mL at this temperature. Calculate the molar mass of naproxen, assuming it remains intact upon dissolution and the density of the solution equals the density of pure benzene. 230 g/mol 176 g/mol 307 g/mol 3.80 × 105 g/mol
Chemistry
1 answer:
sergij07 [2.7K]2 years ago
8 0

Answer:

Molar mass for naproxen is 230 g/mol

Explanation:

Let's apply the colligative property to solve this:

π = M . R . T

4 atm = M . 0.082 l.atm/mol.K . 293K

4 atm / 0.082 mol.K/l.atm . 293K = M → 0.166 mol/L

Molarity are the moles of solute in 1L of solution.

Let's find out the mass of solution, in order to determine the volume (by density) and then, the moles we used.

Solvent + Solute = Solution → 1.138 g + 25 g = 26.138 g

Density of solution = Mass of solution/Volume of solution

Volume of solution = Mass of solute / Density of solution

26.138 g / 0.8787g/mL = 29.7  mL

We convert the volume from mL to L, to multiply by molarity

29.7 mL . 1L/1000 mL = 0.0297L → 0.166 mol/L . 0.0297L = 0.00494 moles

These are the moles, we used in the solution. Let's find out the molar mass → g/mol → 1.138 g / 0.00494 mol = 230.4 g/mol

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Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol
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Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

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(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

Cl=1s^22s^22p^63s^23p^5

Cl^-=1s^22s^22p^63s^23p^6

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

Na=1s^22s^23p^63s^1

Na^+=1s^22s^23p^63s^0

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

Mg=1s^22s^23p^63s^2

Mg^{2+}=1s^22s^23p^63s^0

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

Ca= 1s^22s^22p^63s^23p^64s^2

Ca^{2+}=1s^22s^23p^6^23p^64s^0

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

K= 1s^22s^22p^63s^23p^64s^1

K^{+}=1s^22s^23p^6^23p^64s^0

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5

Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2

Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0

(h) F

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F=1s^22s^22p^5

F^-=1s^22s^22p^6

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