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lord [1]
3 years ago
6

A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi

on some number of amino acids) in water to give 1.70 L of solution. The solution has an osmotic pressure of 4.19 torr at 26 °C. What is the approximate molecular mass of the polypeptide?
Chemistry
1 answer:
faust18 [17]3 years ago
8 0

Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

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

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT

where,

\pi = osmotic pressure of the solution = 4.19 torr

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

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = temperature of the solution = 26^oC=[273+26]K=299K

Putting values in above equation, we get:

4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol

Hence, the molar mass of the polypeptide is 856 g/mol

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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
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Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

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According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

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0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

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The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

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Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

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Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

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0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

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