Studdy AI Logo
Math

QuestionCompare expected and experimental values for ii using osmotic pressure of 0.10M0.10 \mathrm{M} Fe(NH4)2(SO4)2\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} at 25∘C25^{\circ} \mathrm{C}, given 10.8 atm10.8 \mathrm{~atm}.

Studdy Solution

STEP 1

Assumptions1. The molarity of the solution is0.10 M. The compound in the solution is Fe(NH4)(4)\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{}\left(\mathrm{}_{4}\right)_{}
3. The observed osmotic pressure is10.8 atm4. The temperature is 25∘C25^{\circ} \mathrm{C} or 298.15K298.15 \mathrm{K} (Kelvin)
5. We are assuming ideal behavior (i.e., the van 't Hoff factor, ii, is equal to the number of particles the solute dissociates into in solution)

STEP 2

First, let's determine the expected van 't Hoff factor (ii) for the compound Fe(NH4)2(4)2\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{}_{4}\right)_{2}. This compound dissociates into1 Fe2+^{2+} ion,2 NH4+\mathrm{NH}_{4}^{+} ions, and2 42−\mathrm{}_{4}^{2-} ions in solution. So, the expected ii is5.

STEP 3

Now, let's calculate the theoretical osmotic pressure using the formula for osmotic pressureΠ=iMRT\Pi = iMRTwhere Π\Pi is the osmotic pressure, ii is the van 't Hoff factor, MM is the molarity of the solution, RR is the ideal gas constant (0.0821 L atm/mol K), and $$ is the temperature in Kelvin.

STEP 4

Plug in the values for ii, MM, RR, and $$ to calculate the theoretical osmotic pressure.
Π=×0.10 M×0.0821  atm/mol K×298.15 K\Pi = \times0.10 \, \mathrm{M} \times0.0821 \, \mathrm{ \, atm/mol \, K} \times298.15 \, \mathrm{K}

STEP 5

Calculate the theoretical osmotic pressure.
Π=5×0.10 M×0.0821  atm/mol K×298.15 K=12.3 atm\Pi =5 \times0.10 \, \mathrm{M} \times0.0821 \, \mathrm{ \, atm/mol \, K} \times298.15 \, \mathrm{K} =12.3 \, \mathrm{atm}

STEP 6

Now, let's calculate the experimental ii using the observed osmotic pressure and the formula for osmotic pressure. Rearranging the formula to solve for ii gives usi=ΠMRTi = \frac{\Pi}{MRT}

STEP 7

Plug in the values for Π\Pi, MM, RR, and $$ to calculate the experimental $i$.
i=10. atm0.10 M×0.0821  atm/mol K×298.15 Ki = \frac{10. \, \mathrm{atm}}{0.10 \, \mathrm{M} \times0.0821 \, \mathrm{ \, atm/mol \, K} \times298.15 \, \mathrm{K}}

STEP 8

Calculate the experimental ii.
i=10.8 atm0.10 M×0.0821  atm/mol K×298.15 K=4.4i = \frac{10.8 \, \mathrm{atm}}{0.10 \, \mathrm{M} \times0.0821 \, \mathrm{ \, atm/mol \, K} \times298.15 \, \mathrm{K}} =4.4The expected value for ii was5, but the experimental value was found to be4.4. This discrepancy could be due to the compound not fully dissociating in solution or due to interactions between the ions in solution.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord