Question

Mole fraction of urea in its aqueous solution is
1/31.
find mass percentage (% w/w)
of solution ?
(A)5%
(B) 15%
(C) 10%
(D) 20%​

Answer 1

Answer:

Molality=

Weight of the solvent in kg

No. of moles of solute

Given, w/w% of an aqueous solution =20%, i.e., 20 g of solute present in 100 g of solution.

Hence, mass of the solvent present =100−20=80 g

Given, the molecular mass of the solute is 50.

Number of moles of solute (n) =

50

20

=0.4

Now,

Molality(m) =

massofsolvent

No.ofmoles

=

0.08

0.4

=5m

Answer 2

The correct answer is option (c) 10%

GIVEN

Mole fraction of urea = 1/31

TO FIND

Mass percentage of the solution.

SOLUTION

We can simply solve the above problem as follows-

It is given that the solution is an aqueous solution of urea.

this means the solution is made up of water and urea.

We know,

the mole fraction of urea = 1/31

Mole fraction of water = 1-1/31 = 30/31

We know that,

The molar mass of urea = 60g/mol.

The molar mass of water = 18 g/mol.

Let us assume that there is 1 mol of urea solution.

Mass % = (Mass of solute /mass of solution )× 100

Mass of urea = mole fraction × molar mass = 1/31 × 60 = 1.92 grams

Mass of water = mole fraction of water × molar mass of water

Mass of water = 30/31 × 18 = 17.28 grams

$mass\% = \frac{1.92}{1.92 + 17.28} \times 100$

= 10%

Hence, the mass% of an aqueous solution of urea is 10%.

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