At 25° C, at 5% aqueous solution of glucose (molecular weight =180 g mol-1 ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?

Question

At 25° C, at 5% aqueous solution of glucose (molecular weight =180 g mol-1 ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute? (A) 60 (B) 80 (C) 72 (D) 63

Tardigrade Admin · Accepted Answer

Since the two solutions are isotonic, they must have same concentrations in moles/litre. For glucose solution, concentration

= 5 g /100 c m^{3}

(given)

= 50 g / L

∴

\frac{50}{180} = \frac{20}{M}

or

M = 72

(

∵

Molar mass of glucose

= 180 g m o l^{- 1}

)
For unknown substance, concentration

= 2 g /100 c m^{3}

(given)

= 20 g / L = \frac{20}{M} m o l / L

∴

\frac{50}{180} = \frac{20}{M}

or

M = 72

Q. At $25^{\circ} C,$ at 5% aqueous solution of glucose (molecular weight $= 180 g m o l^{- 1}$ ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?

60

80

72

63

98

Q. At 25∘C, at 5% aqueous solution of glucose (molecular weight =180gmol−1 ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?

60

80

72

63

98

Q. At $25^{\circ} C,$ at 5% aqueous solution of glucose (molecular weight $= 180 g m o l^{- 1}$ ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?