Question

An aqueous solution of glucose boils at 100.01 degree celsius. The molal boiling point elevation constant for water is 0.5 K kg per mol. What is the number of glucose molecule in the solution containing 100g of water?

Answer 1

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Answer 2

Answer: $0.012\times 10^{23}$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0=(100-100.01)^0C=0.01^0C$ = Elevation in boiling point

i= vant hoff factor = 1 (for non electrolyte such as glucose)

$K_b$ = boiling point constant = $0.5K/m$

m= molality

$\Delta T_b=i\times K_b\times \frac{\text {number of moles of solute}} {\text{weight of solvent in kg}}$

Weight of solvent (water)= 100 g = 0.1 kg      (1kg=1000g)

number of moles of solute (glucose)= ?

$0.01=1\times 0.5\times \frac{x}{0.1kg}$

$x=0.002moles$

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of glucose contains = $6.023\times 10^{23}$ molecules of glucose

Thus 0.002 moles of glucose contains = $\frac{6.023\times 10^{23}}{1}\times 0.002=0.012\times 10^{23}$ molecules of glucose

The number of glucose molecule in the solution containing 100g of water is $0.012\times 10^{23}$