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Question: An aqueous solution of glucose boils at \[{100.02^0}C\]. What is the number of glucose molecules in ...

An aqueous solution of glucose boils at 100.020C{100.02^0}C. What is the number of glucose molecules in the solution containing 100100 g of water? What will be the osmotic pressure of this glucose solution at 270C{27^0}C(Given Kb{K_b} for water =0.5K.Kg.(mol)1,R=0.082L.atm.(mol.K)1 = 0.5K.Kg.{\left( {mol} \right)^{ - 1}},R = 0.082L.atm.{\left( {mol.K} \right)^{ - 1}}, Avogadro’s constant NA=6.02×1023mol1{N_A} = 6.02 \times {10^{23}}mo{l^{ - 1}})

Explanation

Solution

The molality and molarity will be nearly equal for dilute solutions. The molality can be determined from the number of moles of solute and weight of solution in kilograms. The number of molecules can be obtained number of moles multiplied by Avogadro’s number.

Formula used:
π=CRT\pi = CRT
Where π\pi is osmotic pressure
C is molar concentration or molality
R is ideal gas constant
T is temperature in kelvin

Complete answer:
Given an aqueous solution of glucose boils at 100.020C{100.02^0}C
Temperature = 27+273=300K27 + 273 = 300K
Elevation in boiling point kb=0.5K.Kg.(mol)1{k_{_b}} = 0.5K.Kg.{\left( {mol} \right)^{ - 1}}
Universal gas constant R=0.082L.atm.(mol.K)1R = 0.082L.atm.{\left( {mol.K} \right)^{ - 1}}
Avogadro’s constant NA=6.02×1023mol1{N_A} = 6.02 \times {10^{23}}mo{l^{ - 1}})
The boiling point of water is 1000C{100^0}C
But, given an aqueous solution of glucose boils at 100.020C{100.02^0}C
Thus, elevation in boiling point ΔTb\Delta {T_b}will be
ΔTb=100.020C1000C=0.020C\Delta {T_b} = {100.02^0}C - {100^0}C = {0.02^0}C
But, ΔTb=kb×m\Delta {T_b} = {k_b} \times m
We know the values of ΔTb\Delta {T_b}and kb{k_b}.
Thus, the molality value will be
m=ΔTbkb=0.020.5=0.04m = \dfrac{{\Delta {T_b}}}{{{k_b}}} = \dfrac{{0.02}}{{0.5}} = 0.04
Thus, molality is 0.040.04.
We know that molality is given by number of moles of solute and weight of solvent
Weight of the solvent is 100g=0.1kg100g = 0.1kg
Thus, the number of moles of glucose will be
n=0.040.1=0.004n = \dfrac{{0.04}}{{0.1}} = 0.004
The number of molecules can be obtained by dividing the number of moles of glucose by Avogadro’s number.
Number of molecules of glucose will be
0.004×6.023×1023=2.408×1021molecules0.004 \times 6.023 \times {10^{23}} = 2.408 \times {10^{21}}molecules
Osmotic pressure can be obtained by substituting the values of molality, ideal gas constant and temperature in kelvin.
Thus, osmotic pressure
π=0.04×0.0821×300=0.9852atm\pi = 0.04 \times 0.0821 \times 300 = 0.9852atm

Note:
The molar concentration can also be known as molarity. For dilute solutions like glucose dissolved in water both the terms molality and molarity will be nearly equal. Moles are the units used to express the amount of substance. The number of molecules can be obtained from the number of moles and Avogadro’s number.