How to Calculate Molality for a 0.5% Urea Solution at 0°C?

  • Thread starter Thread starter shaiqbashir
  • Start date Start date
  • Tags Tags
    Molality
AI Thread Summary
To calculate the molality of a 0.5% urea solution at 0°C, first determine the mass of urea in a given volume of solution. A 0.5% solution means there are 0.5 grams of urea per 100 grams of solution. Since the density of the solution is similar to that of water, assume the mass of the solvent (water) is approximately 99.5 grams, or 0.0995 kilograms. Using the formula for molality, which is the number of moles of solute divided by the mass of solvent in kilograms, the molality can be calculated as 0.5 grams of urea (1/60 mol) divided by 0.0995 kg of water. The final molality is approximately 0.0833 mol/kg.
shaiqbashir
Messages
103
Reaction score
0
Molality creating problems for me!

calculate the molality and osmotica pressur of a 0.5% solution of urea (Mol Mass=60) in water at 0 degree celsius.

R-0.0821

now my dear friends!

i have calculated successfuly the osmotic pressure in this question which is

osmotic pressure = 1.86 atm

now the problem is this that how to calculate this molality. i know that

molality = no of moles of the solute/ mass of the solvent in kilograms

but you see that here i don't have any masss of the solute given at all.

Please help me

Thanks in advance
 
Physics news on Phys.org
Such diluted solution has density almost identical with the density of pure water.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top