The mole fraction of glucose (C6H12O6) in an aqueous binary solution is 0.1. The mass percentage of

Q: Why is the sum of mole fractions always 1?

Think of mole fractions like slices of a pizza. If you add up all the slices, you must get the whole pizza back. Mathematically, $\chi_A + \chi_B = \frac{n_A}{n_A+n_B} + \frac{n_B}{n_A+n_B} = \frac{n_A+n_B}{n_A+n_B} = 1$.

Q: How did we get the ratio of moles from the ratio of mole fractions?

Since both mole fractions share the same denominator (total moles $n_A + n_B$), dividing one by the other cancels out the denominator completely, leaving just the ratio of their numerators: $\frac{n_B}{n_A}$.

Q: What are the molar masses of glucose and water?

Water ($H_2O$) has a molar mass of $2(1) + 16 = 18 \text{ g/mol}$. Glucose ($C_6H_{12}O_6$) has a molar mass of $6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g/mol}$.

Q: Why did we calculate the ratio of masses instead of exact masses?

Because mass percentage is an intensive property—it doesn't depend on the total amount of solution you have. Whether you have a drop or a bucket of this solution, the mass ratio and percentage will remain exactly the same.

Q: How do we convert the mass ratio into a mass percentage?

Once you have the ratio $\frac{w_B}{w_A} = \frac{10}{9}$, you can assume the mass of B is $10x$ and the mass of A is $9x$. The total mass is $19x$. The mass percentage of A is simply its part divided by the whole, multiplied by 100: $\frac{9x}{19x} \times 100$.

Question

The mole fraction of glucose $(C_6H_{12}O_6)$ in an aqueous binary solution is $0.1$. The mass percentage of water in it, to the nearest integer, is ......... .

Sigma Percentile · Accepted Answer

The Alchemy of Concentration: From Mole Fractions to Mass Percentages

Have you ever looked at a simple glass of sugar water and wondered about the hidden mathematical symphony happening inside? In chemistry, a solution isn't just a mixture; it's a precise ratio of molecules dancing together. Today, we are going to decode one of the most elegant translations in physical chemistry: converting a mole fraction into a mass percentage.

Imagine you are holding a beaker filled with an aqueous solution of glucose. The problem gives us a single, powerful clue: the mole fraction of glucose is $0.1$. From this tiny seed of information, we will unravel the entire mass composition of the solution. Let's dive in!

### Decoding the Mole Fraction

In any binary solution—a solution with exactly two components—the sum of the mole fractions must always equal a perfect $1$. Think of it like a pizza; no matter how you slice it, the fractions must add up to the whole pie.

Let's designate water as our solvent (Component A) and glucose as our solute (Component B). We are given:
$$ \chi_B = 0.1 $$

Since $\chi_A + \chi_B = 1$, finding the mole fraction of water is a breeze:
$$ \chi_A = 1 - 0.1 = 0.9 $$

Now, what exactly **is** a mole fraction? It is the ratio of the moles of one component to the total moles in the solution. If we take the ratio of the mole fraction of glucose to the mole fraction of water, the "total moles" denominator beautifully cancels out:
$$ \frac{\chi_B}{\chi_A} = \frac{\frac{n_B}{n_A + n_B}}{\frac{n_A}{n_A + n_B}} = \frac{n_B}{n_A} $$

Substituting our values, we get the raw ratio of their molecules:
$$ \frac{n_B}{n_A} = \frac{0.1}{0.9} = \frac{1}{9} $$
This tells us that for every $1$ molecule of glucose, there are exactly $9$ molecules of water!

### The Bridge Between Moles and Mass

While knowing the ratio of molecules is fascinating, our ultimate goal is to find the **mass percentage**. To cross the bridge from the microscopic world of moles to the macroscopic world of mass, we need our trusty toll collector: Molar Mass.

The number of moles ($n$) is simply the given mass ($w$) divided by the molar mass ($M$). 
- For Glucose ($C_6H_{12}O_6$), the molar mass is $M_B = 180 	ext{ g/mol}$.
- For Water ($H_2O$), the molar mass is $M_A = 18 	ext{ g/mol}$.

Let's substitute these into our mole ratio equation:
$$ \frac{n_B}{n_A} = \frac{\frac{w_B}{180}}{\frac{w_A}{18}} = \frac{1}{9} $$

Now, we perform a little algebraic gymnastics. Rearranging the complex fraction gives us:
$$ \frac{w_B}{180} 	imes \frac{18}{w_A} = \frac{1}{9} $$

Notice how beautifully the numbers simplify! $18$ goes into $180$ exactly $10$ times:
$$ \frac{w_B}{10 \cdot w_A} = \frac{1}{9} $$

Multiplying both sides by $10$, we isolate the mass ratio:
$$ \frac{w_B}{w_A} = \frac{10}{9} $$
This is a massive breakthrough. It means that for every $10$ grams of glucose, there are exactly $9$ grams of water in the solution.

### The Final Calculation

We are now standing at the finish line. We need the mass percentage of water. The formula for mass percentage is the mass of the component divided by the total mass, multiplied by $100$.

Using our mass ratio, we can think of the solution in terms of "parts". 
- Mass of water ($w_A$) = $9$ parts
- Mass of glucose ($w_B$) = $10$ parts
- Total mass ($w_A + w_B$) = $9 + 10 = 19$ parts

Plugging this into our percentage formula:
$$ 	ext{Mass \% of Water} = \frac{w_A}{w_A + w_B} 	imes 100 $$
$$ 	ext{Mass \% of Water} = \frac{9}{19} 	imes 100 $$

Calculating this fraction gives us:
$$ 	ext{Mass \% of Water} \approx 47.368\% $$

The question asks us to round to the nearest integer. Looking at the decimal, $47.3$ rounds down to **$47$**.

And there we have it! By simply knowing the fraction of molecules, we were able to deduce that water makes up $47\%$ of the solution's total mass. Chemistry is truly just a puzzle waiting to be solved!

Animated Solution for Chemistry - Basic Concepts in Chemistry: The mole fraction of glucose $(C_{6} H_{12} O_{6})$ in an aqueous binary solution is $0.1$ . The mass percentage of water in it, to the nearest integer, is ......... .

Enter Numerical Value:

Visualized Solution

Understanding the Binary Solution

Mole Fraction Relationship

Ratio of Moles

Converting Moles to Mass

Simplifying the Mass Ratio

Mass Percentage of Water

What's Next?

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Decoding the Mole Fraction

The Bridge Between Moles and Mass

The Final Calculation

Similar Questions

The mole fraction of a solute in a $100$ molal aqueous solution is ......... $\times 1 0^{- 2}$ (Round off to the nearest integer). [Given, atomic masses $H : 1.0 u$ , $O : 16.0 u$ ]

The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm $^{- 3}$ . The ratio of the molecular weights of the solute and solvent, $(\frac{M W _{solute}}{M W _{solvent}})$ , is

The mole fraction of a solvent in aqueous solution of a solute is $0.8$ . The molality (in $mol kg^{- 1}$ ) of the aqueous solution is

A $5.2$ molal aqueous solution of methyl alcohol, $CH_{3} OH$ , is supplied. What is the mole fraction of methyl alcohol in the solution ?

$8 g$ of $NaOH$ is dissolved in $18 g$ of $H_{2} O$ . Mole fraction of $NaOH$ in solution and molality (in $mol kg^{- 1}$ ) of the solution respectively are

If the concentration of glucose ( $C_{6} H_{12} O_{6}$ ) in blood is $0.72 gL^{- 1}$ , the molarity of glucose in blood is ...... $\times 1 0^{- 3} M$ . (Nearest integer) [Given : Atomic mass of $C = 12$ , $H = 1$ , $O = 16 u$ ]

The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is $1.2 g cm^{- 3}$ , the molarity of urea solution is ___ (Given data : Molar masses of urea and water are $60 g mol^{- 1}$ and $18 g mol^{- 1}$ , respectively)

The density of NaOH solution is $1.2 g cm^{- 3}$ . The molality of this solution is ............ m. (Round off to the nearest integer) [Use : Atomic masses : Na=23.0 u, O=16.0 u, H=1.0 u, density of $H_{2} O : 1.0 g cm^{- 3}$ ]

What would be the molality of $20%$ (mass/mass) aqueous solution of KI? (Molar mass of KI = $166 g mol^{- 1}$ )