An aqueous solution of glucose boils at 100.02°C.
What is the number of glucose molecules in the
solution containing 100 g of water? What will be the
osmotic pressure of this glucose solution at 27°C?
(Given : Kb for water = 0.5 K kg mol,
R=0.082 L atm molk, Avogadro's constant,
NA = 6.02 x 1023 mol-')
Answers
TOPIC :- COLLIGATIVE PROPERTIES
FORMULAS
m ( molality) =
∆Tb (Elevation in boiling point ) = Kb × m
π (Osomotic pressure) = CRT
SOLUTIONS :-
Given, glucose solution boils at 100.2 °C.
We know, normal boiling point of water = 100 °C.
Thus, elevation in boiling point = 100.02 - 100. = 0.02 °C.
Now, using formula of ∆Tb
∆Tb = Kb × m
=> 0.02 = 0.5 × m
=> m = 0.02 / 0.5
=> m = 1/25 = 0.04
Now,
weight of solvent = 100 g = 0.1 kg
Hence, m = no. of moles of glucose/0.1
=> 0.04 = n /0.1
=> n = 0.004
Hence, no. of glucose molecules = 0.004 NA ( NA = Avogadro's number).
Now, for dilute solutions molality is approximately equal to molarity.
Hence,
π = CRT
= 0.04 × 0.0821 × 300
= 0.9852 atm.
Hence, osomotic pressure of the solution is 0.9852 atm.
Explanation:
TOPIC :- COLLIGATIVE PROPERTIES
FORMULAS
m ( molality) = \frac{no. \: of \: moles \: of \: solute}{weight \: of \: solvent \: in \: kg}
weightofsolventinkg
no.ofmolesofsolute
∆Tb (Elevation in boiling point ) = Kb × m
π (Osomotic pressure) = CRT
SOLUTIONS :-
Given, glucose solution boils at 100.2 °C.
We know, normal boiling point of water = 100 °C.
Thus, elevation in boiling point = 100.02 - 100. = 0.02 °C.
Now, using formula of ∆Tb
∆Tb = Kb × m
=> 0.02 = 0.5 × m
=> m = 0.02 / 0.5
=> m = 1/25 = 0.04
Now,
weight of solvent = 100 g = 0.1 kg
Hence, m = no. of moles of glucose/0.1
=> 0.04 = n /0.1
=> n = 0.004
Hence, no. of glucose molecules = 0.004 NA ( NA = Avogadro's number).
Now, for dilute solutions molality is approximately equal to molarity.
Hence,
π = CRT
= 0.04 × 0.0821 × 300
= 0.9852 atm.
Hence, osomotic pressure of the solution is 0.9852 atm.