x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05 M NaCl solution. The osmotic pressure of this solution is found to be 4.92 atm at 27°C. Calculate the value of ‘x’ Assume complete dissociation of NaCl and ideal behaviour of this solution

Question

x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05 M NaCl solution. The osmotic pressure of this solution is found to be 4.92 atm at 27°C. Calculate the value of ‘x’ Assume complete dissociation of NaCl and ideal behaviour of this solution (A) 16.52 g (B) 24.132 g (C) 19.959 g (D) 12.35 g

Tardigrade Admin · Accepted Answer

(i)For NaCl π=iCRT=2× 0.05 × 0.0821 × 300=2.463 atm (ii) For unknown compound, π=CRT=(x/200)× 0.0821 ×300=0.1231× atm Total osmotic pressure π=π1+π2 4.92 =2.463+0.1231 X X=19.959 g

$16.52 g$

$24.132 g$

$19.959 g$

$12.35 g$

(i)For NaCl $π = i CRT = 2 \times 0.05 \times 0.0821 \times 300 = 2.463 a t m$
(ii) For unknown compound,
$π = CRT = \frac{x}{200} \times 0.0821 \times 300 = 0.1231 \times a t m$
Total osmotic pressure $π = π_{1} + π_{2}$
$4.92 = 2.463 + 0.1231 X$
$X = 19.959 g$

Q. x g of a non-electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05MNaCl solution. The osmotic pressure of this solution is found to be 4.92atm at 27∘C. Calculate the value of ‘x’ Assume complete dissociation of NaCl and ideal behaviour of this solution

16.52g

24.132g

19.959g

12.35g

(i)For NaCl π=iCRT=2×0.05×0.0821×300=2.463atm (ii) For unknown compound, π=CRT=200x​×0.0821×300=0.1231×atm Total osmotic pressure π=π1​+π2​ 4.92=2.463+0.1231X X=19.959g

$16.52 g$

$24.132 g$

$19.959 g$

$12.35 g$

(i)For NaCl $π = i CRT = 2 \times 0.05 \times 0.0821 \times 300 = 2.463 a t m$
(ii) For unknown compound,
$π = CRT = \frac{x}{200} \times 0.0821 \times 300 = 0.1231 \times a t m$
Total osmotic pressure $π = π_{1} + π_{2}$
$4.92 = 2.463 + 0.1231 X$
$X = 19.959 g$