'V' 1ml of an aqueous urea solution having osmotic pressure equal to 2.4 atm is mixed with 'V' 2ml of another aqueous urea solution having osmotic pressure equal to 4.6 atm at same temperature. What will be value of ( V 1/ V 2) if resulting solution after mixing has osmotic pressure equal to 2.95 atm?

Question

'V' 1ml of an aqueous urea solution having osmotic pressure equal to 2.4 atm is mixed with 'V' 2ml of another aqueous urea solution having osmotic pressure equal to 4.6 atm at same temperature. What will be value of ( V 1/ V 2) if resulting solution after mixing has osmotic pressure equal to 2.95 atm? (A) 1 (B) 1.5 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

π R =(2.4 V 1+4.6 V 2/ V 1+ V 2)=2.95 atm (assume temperature constant) (V1/V2)=3.

1

1.5

2

3

πR​=V1​+V2​2.4V1​+4.6V2​​=2.95 atm (assume temperature constant) V2​V1​​=3.

$π_{R} = \frac{2.4 V _{1} + 4.6 V _{2}}{V _{1} + V _{2}} = 2.95$ atm (assume temperature constant)

$\frac{V _{1}}{V _{2}} = 3$ .