The number of pairs of the solution having the same value of the osmotic pressure from the following is
(Assume 100% ionization)
A. 0.500 M C$$_2$$H$$_5$$OH(aq) and 0.25 M KBr(aq)
B. 0.100 M K$$_4$$[Fe(CN)$$_6$$](aq) and 0.100 M FeSO$$_4$$(NH$$_4$$)$$_2$$SO$$_4$$(aq)
C. 0.05 M K$$_4$$[Fe(CN)$$_6$$](aq) and 0.25 M NaCl(aq)
D. 0.15 M NaCl(aq) and 0.1 M BaCl$$_2$$(aq)
E. 0.02 M KCl.MgCl$$_2$$.6H$$_2$$O(aq) and 0.05 M KCl(aq)

To compare the osmotic pressures of two solutions at the same temperature, we use the relation

$$\pi=iMRT.$$

Since (R) and (T) are constant, two solutions will have the same osmotic pressure if they have the same value of (iM), where (i) is the van't Hoff factor and (M) is the molarity.

For Pair A,

$$C_2H_5OH:\ i=1,\quad iM=1\times0.500=0.50,$$

$$KBr:\ i=2,\quad iM=2\times0.25=0.50.$$

Hence, both solutions have the same osmotic pressure.

For Pair B,

$$K_4[Fe(CN)_6]:\ i=5,\quad iM=5\times0.100=0.50,$$

$$FeSO_4\cdot(NH_4)_2SO_4:\ i=5,\quad iM=5\times0.100=0.50.$$

Hence, both solutions have the same osmotic pressure.

For Pair C,

$$K_4[Fe(CN)_6]:\ iM=5\times0.05=0.25,$$

$$NaCl:\ iM=2\times0.25=0.50.$$

Since the values are different, these solutions do not have the same osmotic pressure.

For Pair D,

$$NaCl:\ iM=2\times0.15=0.30,$$

$$BaCl_2:\ iM=3\times0.10=0.30.$$

Hence, both solutions have the same osmotic pressure.

For Pair E,

$$KCl\cdot MgCl_2\cdot6H_2O:\ i=5,\quad iM=5\times0.02=0.10,$$

$$KCl:\ i=2,\quad iM=2\times0.05=0.10.$$

Hence, both solutions have the same osmotic pressure.

Therefore, Pairs A, B, D, and E have equal osmotic pressures, while Pair C does not. Hence, the number of pairs having the same osmotic pressure is

$$\boxed{4}.$$