V1/T1=V2/T2
524 cm^3/491 K =V2/297K
V2=(524 cm^3*297K)/491 K= 317K
From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.
From the information we have;
Volume of the damp air = 1 L
Pressure of the damp air = 741.0 torr or 0.975 atm
Temperature of the gas = 20 oC + 273 = 293 K
R = 0.082 atm LK-1mol-1
Number of moles = ?
n =PV/RT
n = 0.975 × 1/0.082 × 293
n = 0.041 moles
Volume of water vapor = 1 L
Temperature of water = -10 oC + 273 = 263 K
Pressure of the gas = 607.1 torr or 0.799 atm
R = 0.082 atm LK-1mol-1
n= PV/RT
n = 0.799 × 1/ 0.082 × 263
n = 0.037 moles
Number of moles of water = 0.041 moles - 0.037 moles = 0.004 moles
If 1 mole = 6.02 × 10^23 molecules
0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole
= 2.41 × 10^21 molecules
Learn more: brainly.com/question/2510654
Answer: 
Explanation:
where,
= boiling point of solution = ?
= boiling point of solvent (X) = 
= freezing point constant = 
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte like urea)
= mass of solute (urea) = 29.82 g
= mass of solvent (X) = 500.0 g
= molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
Therefore, the freezing point of solution is 
