3 gm of urea is dissolved in. 45 mg of H2O .the relative lowering in vapour pressure is?

Question

Answer

(P• - p / p•) = n2 / n1p•-p/p• ...is RLVPn2 .no .of moles of soluten1..no. of moles of solventso now ..RLVP = (wt of urea / gram mol. wt of urea.) x( gram mol. wt of H2O / wt of H2O.)urea mol. wt is 60 ..!nd H2O is 18...!they gave 45 mg means 45 x 10^ -3 (milli gram)now do u got that..!?here n1 means no of moles of solvent which is in denominator....so i wrote it reverse..!!!

Answer

Thank you so much Alekhya...I got it...thank you

Answer

20!!

Answer

Calculation of Relative Lowering in Vapour Pressure

Given data:
- Mass of urea (m1) = 3 gm
- Mass of water (m2) = 45 mg
- Molar mass of urea = 60 g/mol
- Vapour pressure of pure water = P°

Step 1: Calculate the number of moles of urea
- Number of moles of urea (n1) = Mass of urea / Molar mass of urea
- n1 = 3 g / 60 g/mol
- n1 = 0.05 mol

Step 2: Calculate the number of moles of water
- Number of moles of water (n2) = Mass of water / Molar mass of water
- n2 = 45 mg / 18 g/mol
- n2 = 0.0025 mol

Step 3: Calculate the total number of moles in the solution
- Total number of moles (n) = n1 + n2
- n = 0.05 mol + 0.0025 mol
- n = 0.0525 mol

Step 4: Calculate the relative lowering in vapour pressure
- According to Raoult's law, the relative lowering in vapour pressure (ΔP/P°) is given by:
- ΔP/P° = x2
- where x2 is the mole fraction of the solute (urea)

Step 5: Calculate the mole fraction of urea
- Mole fraction of urea (x1) = n1 / n
- x1 = 0.05 mol / 0.0525 mol
- x1 = 0.9524

Step 6: Calculate the relative lowering in vapour pressure
- ΔP/P° = x2 = 1 - x1
- ΔP/P° = 1 - 0.9524
- ΔP/P° = 0.0476

Step 7: Expressing the result
- The relative lowering in vapour pressure is 0.0476 or 4.76%.

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