Question

The temperature at which 10% aqueous solution (w/v) of
glucose will exhibit the osmotic pressure of 16.4 atm, is
(R=0.082 dm' atm K-moll)
(b) 180 K (c) 90 K
(d) 300 K (e) 360 K
​

Answer 1

Answer:

d

Explanation:

Answer 2

Solution:

Given values:

Mass by volume percentage of glucose solution $(\frac{w}{V})$ = 10%

$(\frac{w}{V}) = 0.1$

Osmotic pressure, π = 16.4 atm

Molecular weight of glucose = 180g/mol

Osmotic pressure of an aqueous solution is given by formula:

π = CRT

where, C = Concentration of solution

C = $\frac{no.\\ of \\ moles}{Volume}$

number of moles of glucose = $\frac{w}{M}$

C = $\frac{w}{M\\ V}$

where, M = molar mass of glucose

and V = volume of solution in Litre.

By putting the values in osmotic pressure formula we get:

π = $\frac{w}{MV}$ × 1000 × RT

$16.4 = \frac{0.1}{180}$ × 1000 × 0.082 × T

$16.4 = 0.000556$ × 1000 × 0.082 × T

16.4 = 0.04556 × T

T =  $\frac{16.4}{0.0455}$

T = 360 K

10% aqueous solution of glucose will exhibit the osmotic pressure of 16.4 atm at temperature 360 K