Drug has heat of melting 25 K jule/mole. Solubility of drug in water is 0.00082 M at 25 C and 0.00144M at 35 C. Find mole fractions x1 and x2 at each temperature.
One liter of water is 1000/18 = 55.6 moles.
They tell you now many moles of the drug are in a liter.
Get the mole fraction from the ratio
(moles of drug)/(moles total)
I know that way. But what is the difference in mole fraction of drugs and solvent (water) at different temp?
To find the mole fractions (x1 and x2) at each temperature, we can use the formula:
x1 = n1 / (n1 + n2)
x2 = n2 / (n1 + n2)
where x1 and x2 are the mole fractions of the drug at temperatures 25°C and 35°C, respectively, and n1 and n2 are the number of moles of the drug at each temperature.
To calculate the number of moles at each temperature, we need to use the formula:
n = c * V
where n is the number of moles, c is the concentration in moles per liter, and V is the volume in liters.
Given:
Heat of melting of the drug = 25 kJ/mol
Solubility of the drug in water at 25°C = 0.00082 M
Solubility of the drug in water at 35°C = 0.00144 M
Now, let's calculate the number of moles at each temperature:
For 25°C:
n1 = c1 * V1
= 0.00082 mol/L * 1 L
= 0.00082 mol
For 35°C:
n2 = c2 * V2
= 0.00144 mol/L * 1 L
= 0.00144 mol
Now, let's calculate the mole fractions:
For 25°C:
x1 = n1 / (n1 + n2)
= 0.00082 mol / (0.00082 mol + 0.00144 mol)
= 0.00082 / 0.00226
= 0.3628
For 35°C:
x2 = n2 / (n1 + n2)
= 0.00144 mol / (0.00082 mol + 0.00144 mol)
= 0.00144 / 0.00226
= 0.6372
Therefore, the mole fractions at 25°C are approximately x1 = 0.3628 and x2 = 0.6372, while at 35°C, x1 = 0.6372 and x2 = 0.3628.