Osmotic pressure of 40% (W/V) urea solution is 1.64 atm and that of 3.42% (W/V) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of resulting solution is

Question

Osmotic pressure of 40% (W/V) urea solution is 1.64 atm and that of 3.42% (W/V) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of resulting solution is (A) 1.64 atm (B) 2.46 atm (C) 4.10 atm (D) 2.05 atm

Tardigrade Admin · Accepted Answer

π1 V 1+π2 V 2=π R ( V 1+ V 2) 164 × 1+2.46 × 1=π R × 2 π R =(4.10/2) =2.05 atm Here, π R = Osmotic pressure of resulting solution.

Q. Osmotic pressure of 40% (W/V) urea solution is 1.64 atm and that of 3.42% (W/V) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of resulting solution is

1.64 atm

2.46 atm

4.10 atm

2.05 atm

$π_{1} V_{1} + π_{2} V_{2} = π_{R} (V_{1} + V_{2})$

$164 \times 1 + 2.46 \times 1 = π_{R} \times 2$

$π_{R} = \frac{4.10}{2}$

$= 2.05 a t m$

Here, $π_{R} =$ Osmotic pressure of resulting solution.

Q. Osmotic pressure of 40% (W/V) urea solution is 1.64 atm and that of 3.42% (W/V) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of resulting solution is

1.64 atm

2.46 atm

4.10 atm

2.05 atm

π1​V1​+π2​V2​=πR​(V1​+V2​) 164×1+2.46×1=πR​×2 πR​=24.10​ =2.05atm Here, πR​= Osmotic pressure of resulting solution.

$π_{1} V_{1} + π_{2} V_{2} = π_{R} (V_{1} + V_{2})$

$164 \times 1 + 2.46 \times 1 = π_{R} \times 2$

$π_{R} = \frac{4.10}{2}$

$= 2.05 a t m$

Here, $π_{R} =$ Osmotic pressure of resulting solution.