Question

10 g of urea is present in 250 mL of aqueous solution at 27°C. The osmotic pressure of the solution is
a. 8.2 atm
b. 12.4 atm
c. 16.4 atm
d.10.5 atm

please answer if you are sure,​

Answer 1

Answer:

c is the correct answer 16.4 atm

Answer 2

Answer:

⏩ Correct option is C 16.4 atm ⏪

Explanation:

Glucose = 82.4%; Urea = 17.6 %

Let x g glucose and 10−x g urea be present.

The molar masses of glucose and urea are 180 g/mol and 60 g/mol respectively.

The number of moles of glucose = x/180

The number of moles of urea = 10−x /60

Total number of moles of solutes = x/180 + 10−x/60

= x+30−3x/180

= 30−2x/180

= 15−x/90

The molarity of the solution is C= 15−x/90 × 1000/250

= 30−2x/45

The osmotic pressure =Π=CRT

Here, R is the ideal gas constant and T is the absolute temperature.

7.4= 30−2x/45 ×0.08206×300

30−2x=13.53

2x=16.47

x=8.24

10−x=10−8.24=1.76

Percentage of glucose 100x/10 =10x=10×8.24=82.4%

Percentage of urea = 100(10−x)/10

=10(1.76)=17.6%