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Question: If 6g of solute is dissolved in 90g of water. The mole fraction of the solute is? A. \(\dfrac{1}{5...

If 6g of solute is dissolved in 90g of water. The mole fraction of the solute is?
A. 15\dfrac{1}{5}
B. 150\dfrac{1}{50}
C. 151\dfrac{1}{51}
D. 1501\dfrac{1}{501}

Explanation

Solution

The mole fraction of the solute is the number of moles of urea divided by the total number of moles. Hence, first we have to find the number of moles of urea and water using the given mass and the molar mass of these compounds. Dividing the number of moles of urea with the sum of the number of moles of water and urea will give us the answer.

Formulas used: n=WMn = \dfrac{W}{M}
Where nn is the number of moles, WW is the given mass and MM is the molar mass.
xB=nBnA+nB{x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}
Where xB{x_B} is the mole fraction of the solute and nA,nB{n_A},{n_B} represent the number of moles of the solvent and solute respectively.

Complete step by step answer:
Let us first find the number of moles of each component given:
n=WMn = \dfrac{W}{M}
Where nn is the number of moles, WW is the given mass and MM is the molar mass.
For urea (NH2(CO)NH2N{H_2}(CO)N{H_2}), molar mass of each element is: N=14g,H=1g,C=12g,O=16gN = 14g,H = 1g,C = 12g,O = 16g. Hence, its molar mass is: 14+(1×2)+12+16+14+(1×2)=60g14 + (1 \times 2) + 12 + 16 + 14 + (1 \times 2) = 60g
And the given mass is W=6gW = 6g. Substituting these values, we get:
nB=660=110=0.1mol{n_B} = \dfrac{6}{{60}} = \dfrac{1}{{10}} = 0.1mol
Now let us find the number of moles of water
The molar mass of water (H2O{H_2}O) is (1×2)+16=18g(1 \times 2) + 16 = 18g
And the given mass is W=90gW = 90g. Substituting these values, we get:
nA=9018=5mol{n_A} = \dfrac{{90}}{{18}} = 5mol
Mole fraction of the solute is given by the formula:
xB=nBnA+nB{x_B} = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}
Where xB{x_B} is the mole fraction of the solute and nA,nB{n_A},{n_B} represent the number of moles of the solvent and solute respectively.
Substituting the values we got, nB=0.1mol{n_B} = 0.1mol and nA=5mol{n_A} = 5mol, we get:
xB=0.15+0.1=0.15.1\Rightarrow {x_B} = \dfrac{{0.1}}{{5 + 0.1}} = \dfrac{{0.1}}{{5.1}}
On solving this, we get the mole fraction as:
xB=151{x_B} = \dfrac{1}{{51}}

So, the correct answer is Option C .

Note: Mole fraction is one of the methods of expressing the concentration of a component in a mixture. Other methods include molarity, molality, parts per million etc.
Since the number of moles do not change with temperature, mole fraction is a good concentration indicator in temperature dependent reactions.