Molality of an aqueous solution of urea is 444 m Mole fracti

Question

Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is x x 10^-3, Value of x is ________. (Integer answer)

DivineJEE · Accepted Answer

To determine the mole fraction of urea in a solution where the molality is

4.44 \ \mathrm{m}

, we first need to understand the definitions and relationships involved.

Molality (

\mathrm{m}

) is given by:

\mathrm{m} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

Here, the molality is

4.44 \ \mathrm{m}

, which means there are

4.44

moles of urea (solute) per kilogram of water (solvent).

The next step involves calculating the mole fraction. Mole fraction (

X

) of urea is given by:

X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}}

Since we have

4.44

moles of urea, let's determine the moles of water. The molar mass of water (

\mathrm{H_2O}

) is approximately

18 \ \mathrm{g/mol}

. Therefore, the moles of water in

1 \ \mathrm{kg}

(or

1000 \ \mathrm{g}

) of water are:

\text{Moles of water} = \frac{1000 \ \mathrm{g}}{18 \ \mathrm{g/mol}} \approx 55.56 \ \text{moles}

We then substitute these values into the mole fraction formula:

X_{\text{urea}} = \frac{4.44}{4.44 + 55.56}

Simplifying the expression inside the denominator first:

4.44 + 55.56 = 60

So, the mole fraction of urea becomes:

X_{\text{urea}} = \frac{4.44}{60}

Dividing the values gives us:

X_{\text{urea}} \approx 0.074

The problem asks for the mole fraction in the form

x \times 10^{-3}

, so we convert our result to this form:

X_{\text{urea}} = 74 \times 10^{-3}

Therefore, the value of

x

is:

\boxed{74}

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