18 gram glucose (Molar mass = 180) is dissolved in 100 ml of water at 300 K. If R = 3 0.0821 L-atm mol-1 K-1 what is the osmotic pressure of solution ?

Question

18 gram glucose (Molar mass = 180) is dissolved in 100 ml of water at 300 K. If R = 3 0.0821 L-atm mol-1 K-1 what is the osmotic pressure of solution ? (A) 2.463 atm (B) 24.63 atm (C) 8.21 atm (D) 0.821 atm

Tardigrade Admin · Accepted Answer

The various quantities known to us are as follows:

R = 0.0821 L

- atm mol

^{- 1} K^{- 1}

w_{2} = 18

gram

Molar mass

(M_{2}) = 180

T = 300 K

V = 100 m L

To calculate the osmotic pressure of solution, we use the following formula,

π = \frac{w _{2} RT}{M _{2} V} = \frac{18 \times 0.0821 \times 300 \times 1000}{180 \times 100}

= 24.63

atm

Q. $18$ gram glucose (Molar mass $= 180$ ) is dissolved in $100 m l$ of water at $300 K$ . If $R = 3$ $0.0821 L - a t m m o l^{- 1} K^{- 1}$ what is the osmotic pressure of solution ?

$2.463 a t m$

$24.63 a t m$

$8.21 a t m$

$0.821 a t m$

Q. 18 gram glucose (Molar mass =180) is dissolved in 100ml of water at 300K. If R=3 0.0821L−atmmol−1K−1 what is the osmotic pressure of solution ?

2.463atm

24.63atm

8.21atm

0.821atm

The various quantities known to us are as follows: R=0.0821L - atm mol −1K−1 w2​=18 gram Molar mass (M2​)=180 T=300K V=100mL To calculate the osmotic pressure of solution, we use the following formula, π=M2​Vw2​RT​=180×10018×0.0821×300×1000​ =24.63 atm

Q. $18$ gram glucose (Molar mass $= 180$ ) is dissolved in $100 m l$ of water at $300 K$ . If $R = 3$ $0.0821 L - a t m m o l^{- 1} K^{- 1}$ what is the osmotic pressure of solution ?

$2.463 a t m$

$24.63 a t m$

$8.21 a t m$

$0.821 a t m$