Question

The molality of an aqueous solution of sugar(C12H22O11) is 1.62m. Calculate the mole fractions of sugar and water.​

Answer 1

Answer:

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Answer 2

Answer:

The sugar's mole fraction ( solute ) = $0.028$.

The water's mole fraction ( solvent ) = $0.97$.

Explanation:

The molality of the solution = $1.62 molal$

The sugar's mole fraction =?

The water's mole fraction =?

As we know,

• Molality = $\frac{number of moles of solute}{Mass of solvent(kg)}$

Therefore,

• $1.62 molal$ = $\frac{1.62 mole of solute}{1kg of solvent}$

Thus,

• The sugar's number of moles  = $1.62 mol$

• The water's mass = $1kg$ = $1000g$

Therefore,

• The number of moles of solvent ( water ) = $\frac{Mass}{Molar mass}$ = $\frac{1000}{18}$ = $55.55 mol$

Now,

• Given solute's mole fraction ( sugar ) = $\frac{n_{sugar} }{n_{sugar} + n_{water}}$ = $\frac{1.62}{1.62 + 55.55}$ = $\frac{1.62}{57.17}$ = $0.028$.
• Given solvent's mole fraction ( water ) = $\frac{n_{water} }{n_{sugar} + n_{water}}$ = $\frac{55.55}{1.62 + 55.55}$ = $\frac{55.55}{57.17}$ = $0.97$.