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 $= 180 g m o l^{- 1}$ ) in 100 mL of water at $27^{o} C .$ The osmotic pressure of the solution is:
( $R = 0.08206 L$ $atm$ $K^{- 1} m o l^{- 1}$ )

Question

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 m o l^{- 1}

) in 100 mL of water at

27^{o} C .

The osmotic pressure of the solution is:
(

R = 0.08206 L

atm

K^{- 1} m o l^{- 1}

)

Aicharya · Accepted Answer

i factor of glucose and urea are 1 : 1 respectively. Moles of glucose = weight Molecular wieght = 1.8 180 = 10 - 2 Moles of urea = 0.6 60 = 10 - 2 Total mole of solute = 10 - 2 + 10 - 2 = 2 &#215; 10 - 2 Concentration of solution = 2 &#215; 10 - 2 100 &#215; 10 - 3 = 0.2 Osmotic pressure ( &#960; ) = C R T = 0.2 &#215; 0.0821 &#215; 300 = 4.926 &#160; a t m

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60   g   m o l - 1 ) and 1.8…

Solution

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 &#160; g &#160; m o l - 1 ) and 1.8…

Solution

A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g m o l - 1 ) and 1.8…