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Formula to calculate mass percentage of solute is as follows.

Mass percentage of solute = \[\dfrac{\text{Mass of solute}}{\text{Mass of solution}}\times 100\]

In the question it is given that mole fraction of the urea in the aqueous solution is 0.2.

Means in the solution two chemicals are there, urea and water.

We have to calculate mass percentage of solute (urea).

Total number of moles is one.

Total number of moles = number of moles of urea (\[{{n}_{urea}}\]) + number moles of water (\[{{n}_{{{H}_{2}}O}}\])

${{n}_{urea}}+{{n}_{{{H}_{2}}O}}=1$

${{n}_{{{H}_{2}}O}}=1-0.2$

$\text{ = 0}\text{.8}$

We know that, mass percentage of solute = \[\dfrac{\text{mass of solute}}{\text{mass of solution}}\times 100\]

Mass of urea = mole fraction of urea × molecular weight of urea.

=$0.2\times 60$

=12

Mass of solvent = mole fraction water × molecular weight of water.

=$0.8\times 18$

=14.4

Mass of solution = mass of urea + mass of the solvent

= 12+14.4 = 26.4 gm.

Let’s calculate the mass percentage of solute or urea

=$\dfrac{12}{26.4}\times 100$

= 45.45%

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