Question

When $1.80gm$glucose is dissolved in $90gm$of ${{H}_{2}}O$, the mole fraction of glucose is
A.$0.00399$ [AFMC$2000$]
B.$0.00199$
C.$0.0199$
D.$0.998$

Answer 1

Hint: Mole fraction is the number of moles of a component divided by the total number of moles of all the components present in the solution. Therefore we have to find out the mole number of glucose and mole number of water, then put these values in the mole fraction equation.

Formula Used:No. of moles of any component, $n=\dfrac{m}{M}$
$m=$ mass of any component
$M=$The molar mass of that component
Mole fraction of glucose, $x=\dfrac{{{n}_{glu\cos e}}}{{{n}_{glu\cos e}}+{{n}_{water}}}$
Here ${{n}_{glu\cos e}}=$number of moles of glucose
${{n}_{{{H}_{2}}O}}=$ number of moles of water

Complete answer:Here the given solute is glucose and the chemical formula of glucose is ${{C}_{6}}{{H}_{12}}{{O}_{6}}$

The molar mass of glucose,${{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=$($6\times $Atomic weight of carbon)$+$($12\times $Atomic weight of hydrogen)$+$($6\times $Atomic weight of oxygen)
$\therefore {{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=(6\times 12)+(12\times 1)+(6\times 16)=180gm/mol$

Given the mass of glucose,${{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=1.80gm$
$\therefore {{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\dfrac{{{m}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}=\dfrac{1.80gm}{180gm/mol}=0.01mol$

And the given solvent is water and the chemical formula is ${{H}_{2}}O$.
The molar mass of ${{H}_{2}}O$,${{M}_{{{H}_{2}}O}}=$($2\times $ Atomic weight of hydrogen)$+$($1\times $Atomic weight of oxygen)
${{M}_{{{H}_{2}}O}}=(2\times 1)+(1\times 16)=18gm/mol$

Given the mass of water,${{m}_{{{H}_{2}}O}}=90gm$
$\therefore {{n}_{{{H}_{2}}O}}=\dfrac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\dfrac{90gm}{18gm/mol}=5mol$

Now we can determine the mole fraction of glucose,${{X}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}$by taking the ratio of the number of moles of solute to the total sum of the mole number of solute and solvent.

${{X}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\dfrac{{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}{{{n}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}+{{n}_{{{H}_{2}}O}}}$ ………(i)

Putting the above values in equation (i),
${{X}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=\dfrac{0.01}{0.01+5}=1.99\times {{10}^{-3}}$
Or,${{X}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}=0.00199$
Hence the mole fraction of glucose is $0.00199$.

Thus, option (B) is correct.

Note: To approach this type of problem we should take care of the formula of mole fraction, molarity, normality, number of moles, etc. And Atomic weight of chemical elements from the periodic table must be remembered, if it is not possible at least hydrogen to calcium is mandatory.