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…

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}

)

MARKS Content Team · Accepted Answer

Correct answer: (C) 4.92 atm. 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)=2×10^(-2) Concentration of solution =(2×10^(-2))/(100×10^(-3))=0.2 Osmotic pressure (π)=CRT =0.2×0.0821×300=4.926 atm

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:
( $R = 0.08206 L$ $atm$ $K^{- 1} m o l^{- 1}$ )

Solution

Relevant Questions

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 27oC. The osmotic pressure of the solution is:(R=0.08206 L atm K-1 mol-1)

Solution

Relevant Questions

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:
( $R = 0.08206 L$ $atm$ $K^{- 1} m o l^{- 1}$ )