Question

q16

An aqueous solution containing glucose has a vapor pressure of 19.9 torr at 25°C. What would be the vapor pressure of this solution at 45°C? The vapor pressure of pure water is 23.8 torr at 25°C and 71.9 torr at 45°C. Vapor pressure torr If the glucose in the solution were substituted with an equivalent amount (moles) of NaCl, what would be the vapor pressure at 45°C? Vapor pressure torr

Answer 1

Answer) 1) 60.1 torr

2) 120.2 torr

Explanation:

Vapor pressure of solution increases as the temperature increases.

The ratio of vapor pressure of solution to vapor pressure of solvent must remain same ideally.

Therefore,

P_{(Solution,25°C)}:P_{(water,25°C)} =P_{(Solution,45°C)}:P_{(water,45°C)}

19.9 torr/ 23.8 torr =P(Solution,45°C)/71.9 torr

P_{(Solution,45°C)}=[(19.9 × 71.9)/23.8] torr

P_{(Solution,45°C)}= 60.1 torr

Vapor pressure formula:

π = i MRT

Here, π is vapor pressure, i is the ratio of number of moles of particles in solution to the number of moles of solute in solution, M is the concentration/molarity, R is gas constant and T is the temperature.

R is a constant, T= 45°C therefore constant , M is concentration is constant for both glucose and NaCl at 45°C because number of moles are equivalent.

Hence π is only directly proportional to i.

i for glucose is 1 because it dissolves im water but doesn't dissociate.

i for NaCl is 2 because 1 mol NaCl dissociate into 1 mol Na+ and 1 mol Cl-.

Ratio of NaCl to gluose vapor pressure .

π_(NaCl,45°C)/π_{(glucose,45°C)} = i_{​​​​​NaCl}/i_{​​​​​​glucose}

π_(NaCl,45°C)/(60.1 torr) = 2/1

π_(NaCl,45°C)= 2× 60.1 torr = 120.2 torr