Question

A solution is prepared by dissolving 28.8 gg of glucose (C6H12O6)(C6H12O6) in 355 gg of water. The final volume of the solution is 384 mLmL . For this solution, calculate each of the following.

a) molarity

b) molality

c) percent by mass

d) mole fraction

e) mole percent

Answer 1

sol:-

• 1 kg = 1000 g
• 1 L = 1000 mL
• atomic mass of C = 12 g
• atomic mass of O = 16 g
• atomic mass of H = 1 g
• molecular mass of C₆H₁₂O₆ = [(6 *12) + (12 * 1) + (6 * 16)] g . =[72 + 12 + 96 ] g. = 180 g
• molecular mass of H₂O = [ (2 * 1) + 16] g. = 18 g

data provided in the question is :-

• amount of glucose( C₆H₁₂O₆) = 28.8 g
• amount of water (H₂O) = 355 g = 0.355 kg
• final volume (total volume) = 384 mL = 0.387 L
• mass of solution = (28.8 + 355 ) g = 383.8 g
• solute = glucuse(C₆H₁₂O₆)
• solvent = water (H₂O)
• solution = solute + solvent

Formula used :-

• Molarity = (moles of solute) / (volume of solution in Liters)
• molality = ( moles of solute) / ( mass of solvent in kg)
• mass percentage = {(mass of solute or solvent) / (total mass of solution} * 100
• mole fraction = moles / total mole
• mole percentage = (mole fraction) * 100
• mole = (mass of compound) / (molecular mass of compound)

Calculation :-

putt all the value in proper fromula:-

mole of H₂O = (mass of H₂O )/ (molecular mass of H₂O)

= 355 / 18

= 19.72

moles of glucose(C₆H₁₂O₆) = (mass of C₆H₁₂O₆) / (molecular mass of C₆H₁₂O₆)

= 28.8 / 180

= 0.16

Total mole = (mole of H₂O) + (mole of C₆H₁₂O₆)

= (19.72 + 0.16 ) mole

= 19.88 mole

Hence from the formula and value :-

(a) :- molarity = mole of solute / volume of solution in Litres

=( 0.16 / 0.387 ) mole/ L

= 0.4134 mole / L

molarity = 0.4134 mole / L

(b) :- molality = (moles of solute) /(mass of solvent in kg)

=( 0.16/ 0.355 ) mole/kg

= 0.4507 mole/kg

molality = 0.4507 mole/kg

(c):- mass % of H₂O = {( mass of H₂O)/ (tatal mass of solution)} * 100.

= (355 g / 383.8 g) * 100

= 0.924960 * 100

= 92.4960 %

mass % of H₂O = 92.4960 %

mass % of C₆H₁₂O₆ = (100 - 92.4960) %

= 7.504 %

mass % of C₆H₁₂O₆ = 7.504 %

(d):- mole fraction of H₂O = (mole of H₂O) / (total mole)

= 19.72 / 19.88

= 0.99195

mole fraction of H₂O = 0.99195

mole fraction of C₆H₁₂O₆ = (mole of C₆H₁₂O₆) / (total mole)

= 0.16 / 19.88

= 0.00805

mole fraction of C₆H₁₂O₆ = 0.00805

(e) :- mole % of H₂O = (mole fraction of H₂O) * 100

= 0.99195 * 100

= 99.195 %

mole % of H₂O = 99.195 %

mole % of C₆H₁₂O₆ = (mole fraction of C₆H₁₂O₆) * 100

= 0.00805 * 100

= 0.805 %  

Mole % of C₆H₁₂O₆ = 0.805 %