Question

How many grams of water must be used to dissolve 100.0 grams of sucrose to prepare a 0.020 mole fraction of sucrose(C12H22O11)in the solution?​

Answer 1

Answer:

5000 gram water to dissolve

Answer 2

Answer:

The amount of water required to dissolve sucrose (C₁₂H₂₂O₁₁) is equal to 258 grams.

Explanation:

We have given, mole fraction of sucrose in the solution $X_{s}= 0.020$

The mass of sucrose will be dissolved = 100g

The molecular mass of sucrose (C₁₂H₂₂O₁₁) = 342g/mol

The number of moles of C₁₂H₂₂O₁₁ sucrose, $N_{s}$ = 100/342 = 0.292mol

Mole fraction of sucrose = moles of sucrose/(moles of sucrose + moles of water)

$X_{s}=\frac{N_{s}}{N_{s}+N_{w}}$

$0.020=\frac{0.292}{0.292+N_{w}}$

$0.292+N_{w}= \frac{0.292}{0.020}$

$0.292+N_{w}=14.62$

$N_{w} = 14.33mol$

The number of moles of water = 14.33 moles

The molecular mass of water = 18g/mol

Mass of water = moles of H₂O × molar mass of H₂O

Mass of water = 14.33 ×18 = 258g

Therefore, the mass of water equals to 258g, is required to dissolve 100g of sucrose.