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liq [111]
2 years ago
8

The osmotic pressure exerted by a solution is equal to the molarity multiplied by the absolute temperature and the gas constant

R. Suppose the osmotic pressure of a certain solution is measured to be 20.atm at an absolute temperature of 321.K. Write an equation that will let you calculate the molarity c of this solution. Your equation should contain only symbols. Be sure you define each symbol other than R.
Chemistry
1 answer:
Elena-2011 [213]2 years ago
4 0

Answer:

20 atm / (R . 321 K . i) = M

Explanation:

Osmotic pressure is a colligative property which is determined by the solute.

Solutes over solutions, can increase the osmotic pressure which is calculated like this:

P = M . R . T

where P is pressure, M is molarity, R the Universal Constant Gases and T, the absolute Temperature. We also have to consider the Van't Hoff factor, i, numbers of ions dissolved. Ionic salts determine more osmotic pressure than organic compounds, where i = 1. Then, the complete formula is:

π =  M . R . T . i

To determine molarity, osmotic pressure divide all the terms:

20 atm / (R . 321 K . i) = M

R = 0.082 L.atm/mol.K

If you see all the units in R, they will be cancelled, so finally the answer will be mol/L, which is molarity. The i term has no unit!.

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You can practice converting between the mass of a solution and mass of solute when the mass percent concentration of a solution
Ostrovityanka [42]

Answer:

145 grams of a 0.480 % KCN solution contains 697 mg of KCN.

Explanation:

w/w % : The percentage mass or fraction is mass of the of solute present in 100 grams of the solution.

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

given =  w/w% = 0.480 %

Mass of solute that KCN = 697 mg = 0.697 g

(1 mg = 0.001 g)

Mass of the solution = ? = x

0.480\%=\frac{0.697 g}{x}\times 100

x=\frac{0.697 g}{0.480}\times 100

x = 145.21 g ≈ 145 g

145 grams of a 0.480 % KCN solution contains 697 mg of KCN.

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