Question

2. an aqueous solution of a certain organic compound has a density of 1.063 gml-1, an osmotic pressure of 12.16 atm at 250 c and a freezing point of -1.030 c. what is the molar mass of the compound? plzzz it's urgent plzzz plzzz plzzzz plzzzzzz plzzzzz​

Answer 1

here is the ans .

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Answer 2

Answer:

The molar mass of the compound will be 334 gmol⁻¹.

Calculation:

In order to calculate the molar mass of the compound, we will need to go through various steps using different formulas. The main ones are written below:

1. Osmotic pressure → π = M × R × T
2. Freezing point → ΔTf = Kf × m
3. Molality → $\frac{1000W2}{M2W1}$

In the question, the data given to us is:

1. Osmotic pressure → 12.16 atm
2. Density → 1.063 gml⁻¹
3. Temperature → 25°C = 298.15 K
4. Freezing point → -1.03 °C

- First, find out ΔTf:

ΔTf = ΔT°f - Tf

0° - (-1.03)

ΔTf = 1.03°C

- Second, find the molality with the freezing point formula.

ΔTf = Kf × m

The experimental value of Kf of water is 1.86°Ckg/mol.

1.03 = 1.86 × m

molality = 0.554 m

Third, using the osmotic pressure formula we find the molarity.

π = M × R × T

12.16 = M × 0.082 × 298.15

M = $\frac{12.16}{0.082*298.15}$

Molarity = 0.479M

Fourth, find the mass of solvent using the $\frac{Molarity}{Density}$ formula, followed by mass of solution and mass of solute.

Mass of solvent = $\frac{0.497}{0.554}$

Mass of solvent = 897 g

Mass of solution = 1.063 × 1000 = 1063 g

Mass of solute = 1063 - 987 = 166 g

At last, we use the molality formula to calculate the molar mass.

m= $\frac{1000W2}{M2W1}$ where

1. W₂, mass of solute = 166 g
2. M₂, mass of the compound
3. m= molality = 0.554 M
4. W₁, mass of solvent = 897 g

0.554 = $\frac{1000*166}{M2*897}$

M₂ = $\frac{1000*166}{0.55*897}$

M₂ = 334 gmol⁻¹

Conclusion:

The molar mass is 334 gmol⁻¹.