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The following solutions were prepared by dissolving 10 g of glucose ( C 6 H 12 O 6 ) (C_6H_{12}O_6) ( C 6 โ H 12 โ O 6 โ ) in 250 ml of water ( P 1 ) , 10 g (P_1), 10 g ( P 1 โ ) , 10 g of urea ( C H 4 N 2 O ) (CH_4N_2O) ( C H 4 โ N 2 โ O ) in 250 ml of water ( P 2 ) (P_2) ( P 2 โ ) and 10 g of sucrose ( C 12 H 22 O 11 ) (C_{12}H_{22}O_{11}) ( C 12 โ H 22 โ O 11 โ ) in 250 ml of water ( P 3 ) . (P_3). ( P 3 โ ) . The right option for the decreasing order of osmotic pressure of these solutions is:
medium
Solutions
2021
chemistry
P 2 > P 1 > P 3 P_2 > P_1 > P_3 P 2 โ > P 1 โ > P 3 โ
Explanation To determine the order of osmotic pressure for the given solutions, we need to use the formula for osmotic pressure: ฮ = i โ
C โ
R โ
T \\
\Pi = i \cdot C \cdot R \cdot T \\
ฮ = i โ
C โ
R โ
T Where: \\
โข ฮ \Pi ฮ is the osmotic pressure. \\
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P 1 > P 2 > P 3 P_1 > P_2 > P_3 P 1 โ > P 2 โ > P 3 โ
P 2 > P 3 > P 1 P_2 > P_3 > P_1 P 2 โ > P 3 โ > P 1 โ
P 3 > P 1 > P 2 P_3 > P_1 > P_2 P 3 โ > P 1 โ > P 2 โ
โข
is the van 't Hoff factor (number of particles the solute dissociates into).
โข
is the molar concentration of the solution.
โข
is the ideal gas constant.
โข
is the temperature in Kelvin.
Since all solutions are prepared in the same volume of water (250 ml), and assuming the temperature is constant,
the osmotic pressure is directly proportional to the product of the van 't Hoff factor and the molar concentration:
ฮ โ i โ
C \\
\Pi \propto i \cdot C \\
ฮ โ i โ
C Let's calculate the molar concentration for each solution:
1. For glucose
( C 6 H 12 O 6 ) : (C_6H_{12}O_6): \\
( C 6 โ H 12 โ O 6 โ ) : Molar mass of glucose
= 180 โ g / m o l = 180 \, \mathrm{g/mol} \\
= 180 g/mol Moles of glucose
= 10 โ g 180 โ g / m o l = 1 18 โ m o l = \frac{10 \, \mathrm{g}}{180 \, \mathrm{g/mol}} = \frac{1}{18} \, \mathrm{mol} \\
= 180 g/mol 10 g โ = 18 1 โ mol Concentration
C 1 = 1 18 โ m o l 0.250 โ L = 1 4.5 โ m o l / L C_1 = \frac{\frac{1}{18} \, \mathrm{mol}}{0.250 \, \mathrm{L}} = \frac{1}{4.5} \, \mathrm{mol/L} \\
C 1 โ = 0.250 L 18 1 โ mol โ = 4.5 1 โ mol/L Van 't Hoff factor
(glucose does not dissociate)
2. For urea
( C H 4 N 2 O ) : (CH_4N_2O): \\
( C H 4 โ N 2 โ O ) : Molar mass of urea
= 60 โ g / m o l = 60 \, \mathrm{g/mol} \\
= 60 g/mol Moles of urea
= 10 โ g 60 โ g / m o l = 1 6 โ m o l = \frac{10 \, \mathrm{g}}{60 \, \mathrm{g/mol}} = \frac{1}{6} \, \mathrm{mol} \\
= 60 g/mol 10 g โ = 6 1 โ mol Concentration
C 2 = 1 6 โ m o l 0.250 โ L = 2 3 โ m o l / L C_2 = \frac{\frac{1}{6} \, \mathrm{mol}}{0.250 \, \mathrm{L}} = \frac{2}{3} \, \mathrm{mol/L} \\
C 2 โ = 0.250 L 6 1 โ mol โ = 3 2 โ mol/L Van 't Hoff factor
(urea does not dissociate)
3. For sucrose
( C 12 H 22 O 11 ) : (C_{12}H_{22}O_{11}): \\
( C 12 โ H 22 โ O 11 โ ) : Molar mass of sucrose
= 342 โ g / m o l = 342 \, \mathrm{g/mol} \\
= 342 g/mol Moles of sucrose
= 10 โ g 342 โ g / m o l = 5 171 โ m o l = \frac{10 \, \mathrm{g}}{342 \, \mathrm{g/mol}} = \frac{5}{171} \, \mathrm{mol} \\
= 342 g/mol 10 g โ = 171 5 โ mol Concentration
C 3 = 5 171 โ m o l 0.250 โ L = 20 171 โ m o l / L C_3 = \frac{\frac{5}{171} \, \mathrm{mol}}{0.250 \, \mathrm{L}} = \frac{20}{171} \, \mathrm{mol/L} \\
C 3 โ = 0.250 L 171 5 โ mol โ = 171 20 โ mol/L Van 't Hoff factor
(sucrose does not dissociate)
Now, calculate the osmotic pressure for each solution:
ฮ 1 โ i 1 โ
C 1 = 1 โ
1 4.5 = 1 4.5 ฮ 2 โ i 2 โ
C 2 = 1 โ
2 3 = 2 3 ฮ 3 โ i 3 โ
C 3 = 1 โ
20 171 = 20 171 \\
\Pi_1 \propto i_1 \cdot C_1 = 1 \cdot \frac{1}{4.5} = \frac{1}{4.5} \\
\Pi_2 \propto i_2 \cdot C_2 = 1 \cdot \frac{2}{3} = \frac{2}{3} \\
\Pi_3 \propto i_3 \cdot C_3 = 1 \cdot \frac{20}{171} = \frac{20}{171} \\
ฮ 1 โ โ i 1 โ โ
C 1 โ = 1 โ
4.5 1 โ = 4.5 1 โ ฮ 2 โ โ i 2 โ โ
C 2 โ = 1 โ
3 2 โ = 3 2 โ ฮ 3 โ โ i 3 โ โ
C 3 โ = 1 โ
171 20 โ = 171 20 โ Comparing the values:
2 3 > 1 4.5 > 20 171 \\
\frac{2}{3} > \frac{1}{4.5} > \frac{20}{171} \\
3 2 โ > 4.5 1 โ > 171 20 โ Thus, the decreasing order of osmotic pressure is:
P 2 > P 1 > P 3 \\
P_2 > P_1 > P_3 \\
P 2 โ > P 1 โ > P 3 โ Therefore, the correct option is Option 1.