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Romashka-Z-Leto [24]
4 years ago
13

The nonvolatile, nonelectrolyte glucose, C6H12O6 (180.2 g/mol), is soluble in water H2O. Calculate the osmotic pressure (in atm)

generated when 11.9 grams of glucose are dissolved in 217 mL of a water solution at 298 K.
Chemistry
1 answer:
Lilit [14]4 years ago
4 0

<u>Answer:</u> The osmotic pressure of the solution is 7.44 atm

<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 = ?

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

m_{solute} = given mass of glucose = 11.9 g

M_{solute} = molar mass of glucose = 180.2 g/mol

V_{solution} = Volume of solution = 217 mL

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

T = temperature of the solution = 298 K

Putting values in above equation, we get:

\pi=1\times \frac{11.9\times 1000}{180.2\times 217}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=7.44atm

Hence, the osmotic pressure of the solution is 7.44 atm

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<em>A solution is prepared at 25 °C that is initially 0.14 M in diethylamine, a weak base with Kb = 1.3 × 10⁻³, and 0.20 M in diethylammonium chloride. Calculate the pH of the solution. Round your answer to 2 decimal places.</em>

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