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dalvyx [7]
3 years ago
7

The nonvolatile, nonelectrolyte testosterone, C19H28O2 (288.40 g/mol), is soluble in benzene C6H6. Calculate the osmotic pressur

e generated when 12.4 grams of testosterone are dissolved in 168 ml of a benzene solution at 298 K.

Chemistry
2 answers:
egoroff_w [7]3 years ago
7 0

Answer: 6.26atm

Explanation:Please see attachment for explanation

BaLLatris [955]3 years ago
3 0

Answer:

The osmotic pressure is 6.26 atm

Explanation:

Step 1: Data given

Mass of testosterone = 12.4 grams

Volume of benzene = 168 mL

Temperature = 298 Kelvin

Step 2: Calculate moles of testosterone

Moles testosterone = mass / molar mass

Moles testosterone = 12.4 grams / 288.42 g/mol

Moles testosterone = 0.0430 moles

Step 3: Calculate molarity

Molarity = moles / volume

Molarity = 0.0430 moles / 0.168 L

Molarity = 0.256 M

Step 4 : Calculate the osmotic pressure

π = iMRT

⇒ with i = The Van't hoff factor = 1 (since testosterone is nonelectrolyte)

⇒ with M = the molair concentration = 0.256 M

⇒ with R = the gas constant = 0.08206 L*atm/k*mol

⇒ with T = the temperature = 298 K

π = 1*0.256*0.08206*298

π = 6.26 atm

The osmotic pressure is 6.26 atm

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A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 295 mL of solution. How many moles o
Masja [62]

Answer:

(a) Moles of ammonium chloride = 0.243 moles

(b) Molarity_{ammonium\ chloride}=0.824\ M

(c) 60.68 mL

Explanation:

(a) Mass of ammonium chloride = 13.0 g

Molar mass of ammonium chloride = 53.491 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{13.0\ g}{53.491\ g/mol}

<u>Moles of ammonium chloride = 0.243 moles</u>

(b) Moles of ammonium chloride = 0.243 moles

Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{ammonium\ chloride}=\frac{0.243}{0.295}

Molarity_{ammonium\ chloride}=0.824\ M

(c) Moles of ammonium chloride = 0.0500 moles

Volume = ?

Molarity = 0.824 M

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

0.824\ M=\frac{0.0500}{Volume}

<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>

6 0
3 years ago
How much NaOH (in grams) is needed to prepare 463 mL of solution with a pH of 10.020?
tino4ka555 [31]

Answer: 1.94\times 10^{-3}g of NaOH

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

Putting in the values:

10.020=-\log[H^+]

[H^+]=9.55\times 10^{-11}

[H^+][OH^-]=10^{-14}

[OH^-]=\frac{10^{-14}}{9.55\times 10^{-11}}=1.05\times 10^{-4}M

NaOH\rightarrow Na^++OH^-

Molarity=\frac{moles\times 1000}{\text {Volume in ml}}

1.05\times 10^{-4}M=\frac{moles\times 1000}{463ml}

moles = 4.86\times 10^{-5}

Mass of NaOH=moles\times {\text {Molar mass}}=4.86\times 10^{-5}\times 40=1.94\times 10^{-3}g

Thus 1.94\times 10^{-3}g of NaOH is needed to prepare 463 mL of solution with a pH of 10.020

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5 drops of 0.15 M Ki added to<br>40 drops of Na2S2O3<br>What is the final concentration of ki?​
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Answer:

\boxed{\text{0.017 mol/L}}

Explanation:

Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

c_{1}V_{1} = c_{2}V_{2}

Data:

c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops

V(Na₂S₂O₃) = 40 drops

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V₂ = 5 + 40 = 45 drops

(b) Calculate the concentration

0.15 × 5 = c₂ × 45

0.75 = 45c₂

c_{2} = \dfrac{0.75 }{45} = \boxed{\textbf{0.017 mol/L}}

3 0
3 years ago
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