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valentina_108 [34]
3 years ago
6

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution

is measured to be 0.0861 atm at 25.0 °C. Calculate the molar mass of the protein. Round your answer to 3 significant digits. mol x 6 ?
Chemistry
1 answer:
ch4aika [34]3 years ago
4 0

<u>Answer:</u> The molecular weight of protein is 1.14\times 10^2g/mol

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT

where,

\pi = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of protein = 400 mg = 0.4 g   (Conversion factor:  1 g = 1000 mg)

M_{solute} = molar mass of protein = ?

V_{solution} = Volume of solution = 5.00 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol

Hence, the molecular weight of protein is 1.14\times 10^2g/mol

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Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

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We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

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We can also think [OH⁻] = Kw / [H⁺] so:

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[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

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(-b +- √(b² - 4ac) / (2a)

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A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

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