A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol-1) and 1.8 g of glucose (molar mass = 180 g mol-1) in 100 mL of water at 27°C. The osmotic pressure of the solution is: (R = 0.08206 L atm K-1 mol-1 )

Question

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol-1) and 1.8 g of glucose (molar mass = 180 g mol-1) in 100 mL of water at 27°C. The osmotic pressure of the solution is: (R = 0.08206 L atm K-1 mol-1 )  (A) 4.92 atm (B) 1.64 atm (C) 2.46 atm (D) 8.2 atm

Tardigrade Admin · Accepted Answer

Π = (((0.6/60) + (1.8/180))/0.1) ×0.08206 ×30 Π =4.9236 textatm

Q. A solution is prepared by dissolving $0.6 g$ of urea (molar mass = $60 g m o l^{- 1})$ and $1.8 g$ of glucose (molar mass = $180 g$ mol $- 1$ ) in $100 m L$ of water at $2 7^{\circ} C$ . The osmotic pressure of the solution is :
$(R = 0.08206 L a t m K^{- 1} m o l^{- 1})$

4.92 atm

1.64 atm

2.46 atm

8.2 atm

$Π = \frac{( \frac{0.6}{60} + \frac{1.8}{180} )}{0.1} \times 0.08206 \times 30$
$Π = 4.9236 atm$

Q. A solution is prepared by dissolving 0.6g of urea (molar mass = 60gmol−1) and 1.8g of glucose (molar mass = 180g mol−1) in 100mL of water at 27∘C. The osmotic pressure of the solution is : (R=0.08206LatmK−1mol−1)