Question

the boiling point of a glucose solution containing 12 gram of glucose in hundred gram of water is hundred. 34 degree Celsius calculate the molar elevation constant of water boiling point of water is hundred degree Celsius​

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Secondary School Chemistry 10+5 pts

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?

by Rishinisha8979 08.08.2018

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aditimee06

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KobenhavnExpert

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}

Explanation:

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