JEE Challenger
More from Solutions
ChemMedAdv 2023 • Paper 2

Osmotic Pressure of Mixed Urea Solutions Calculation

50 mL50\text{ mL} of 0.2 molal0.2\text{ molal} urea solution (density =1.012 g mL−1= 1.012\text{ g mL}^{-1} at 300 K300\text{ K}) is mixed with 250 mL250\text{ mL} of a solution containing 0.06 g0.06\text{ g} of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at 300 K300\text{ K} is _____.

[Use: Molar mass of urea =60 g mol−1= 60\text{ g mol}^{-1}; gas constant, R=62 L Torr K−1 mol−1\text{R} = 62\text{ L Torr K}^{-1}\text{ mol}^{-1};
Assume, ΔmixH=0\Delta_{\text{mix}}\text{H} = 0, ΔmixV=0\Delta_{\text{mix}}\text{V} = 0]

Official Numerical Answer682

Topics & Concepts

SolutionsConcentration Terms

Step-by-Step Solution

To find the osmotic pressure of the resulting solution, we need to calculate the total number of moles of urea and the total volume of the solution after mixing.

Step 1: Calculate moles of urea in Solution 1

For Solution 1:

  • Volume of solution, V1=50 mLV_1 = 50\text{ mL}
  • Density of solution, d1=1.012 g mL−1d_1 = 1.012\text{ g mL}^{-1}

The total mass of Solution 1 is: Mass of solution 1=V1×d1=50 mL×1.012 g mL−1=50.6 g\text{Mass of solution } 1 = V_1 \times d_1 = 50\text{ mL} \times 1.012\text{ g mL}^{-1} = 50.6\text{ g}

Let n1n_1 be the number of moles of urea in Solution 1.

  • Mass of urea in Solution 1 = n1×Murea=60n1 gn_1 \times M_{\text{urea}} = 60 n_1\text{ g}
  • Mass of solvent in Solution 1 = (50.6−60n1) g=50.6−60n11000 kg(50.6 - 60 n_1)\text{ g} = \frac{50.6 - 60 n_1}{1000}\text{ kg}

Using the given molality (m=0.2 mol kg−1m = 0.2\text{ mol kg}^{-1}): m=n1Mass of solvent in kgm = \frac{n_1}{\text{Mass of solvent in kg}} 0.2=n150.6−60n110000.2 = \frac{n_1}{\frac{50.6 - 60 n_1}{1000}} 0.2×(50.6−60n1)=1000n10.2 \times (50.6 - 60 n_1) = 1000 n_1 10.12−12n1=1000n110.12 - 12 n_1 = 1000 n_1 1012n1=10.12  ⟹  n1=0.01 mol1012 n_1 = 10.12 \implies n_1 = 0.01\text{ mol}

Step 2: Calculate moles of urea in Solution 2

For Solution 2:

  • Mass of urea, w2=0.06 gw_2 = 0.06\text{ g}
  • Molar mass of urea, Murea=60 g mol−1M_{\text{urea}} = 60\text{ g mol}^{-1}

Moles of urea in Solution 2 (n2)=0.06 g60 g mol−1=0.001 mol\text{Moles of urea in Solution 2 } (n_2) = \frac{0.06\text{ g}}{60\text{ g mol}^{-1}} = 0.001\text{ mol}

Step 3: Calculate total moles and total volume

  • Total moles of urea (ntotaln_{\text{total}}): ntotal=n1+n2=0.01 mol+0.001 mol=0.011 moln_{\text{total}} = n_1 + n_2 = 0.01\text{ mol} + 0.001\text{ mol} = 0.011\text{ mol}

  • Since ΔmixV=0\Delta_{\text{mix}}V = 0, the total volume of the resulting solution (VtotalV_{\text{total}}) is: Vtotal=V1+V2=50 mL+250 mL=300 mL=0.3 LV_{\text{total}} = V_1 + V_2 = 50\text{ mL} + 250\text{ mL} = 300\text{ mL} = 0.3\text{ L}

Step 4: Calculate the osmotic pressure (Π\Pi)

The osmotic pressure is given by the formula: Π=CRT=(ntotalVtotal)RT\Pi = C R T = \left(\frac{n_{\text{total}}}{V_{\text{total}}}\right) R T

Given:

  • R=62 L Torr K−1 mol−1R = 62\text{ L Torr K}^{-1}\text{ mol}^{-1}
  • T=300 KT = 300\text{ K}

Substituting the values: Π=(0.011 mol0.3 L)×62 L Torr K−1 mol−1×300 K\Pi = \left( \frac{0.011\text{ mol}}{0.3\text{ L}} \right) \times 62\text{ L Torr K}^{-1}\text{ mol}^{-1} \times 300\text{ K} Π=0.011×62×1000\Pi = 0.011 \times 62 \times 1000 Π=11×62=682 Torr\Pi = 11 \times 62 = 682\text{ Torr}

Final Answer:

The osmotic pressure of the resulting solution at 300 K300\text{ K} is 682.

PreviousNext