solution of -glucose in wa

$250 \mathrm{~g}$ solution of $\mathrm{D}$ -glucose in water contains $10.8 \%$ of carbon by weight. The molality of the solution is nearest to
(Given: Atomic Weights are, $\mathrm{H}, 1 \mathrm{u} ; \mathrm{C}, 12 \mathrm{u} ; \mathrm{O}, 16 \mathrm{u}$ )
Option: 1
$1.03$

Option: 2
$2.06$

Option: 3
$3.09$

Option: 4
$5.40$

Answers (1)

We need to think of the reverse solution

$\text{molality}\rightarrow \text { mole of solute } \rightarrow \text { mass of } \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
                    $\rightarrow \text{weight of solvent} \rightarrow \left [ 250g-\text{mass of} \, \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \right ]$

$\mathrm{mass\: of\: \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \rightarrow moles \: of \: C_{6}\: present}$
                                                        $\mathrm{\downarrow}$
                                           $\text{ mass of carbon in solution.}$

Now, 10.8 % of carbon by weight present in a 250 g solution
$\mathrm{mass \: of \: carbon =\frac{250}{100} \times 10.8=27 \mathrm{~g} }$

$\mathrm{Hence , moles\: of\: 6 C \left(C_6\right) present =\frac{27}{12 \times 6}=0.375 \, moles }$

$\mathrm{mole\: of \: \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6=0.375 \: moles -(1) }$
$\mathrm{mass \: of\: \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6=0.375 \times 180=67.5 \mathrm{~g} }$
$\mathrm{So, mass\: of \: solvent =250 \mathrm{~g}-67.5 \mathrm{~g}=182.5 g}$
$\mathrm{mass \: of \: solvent =182.5 \times 10^{-3} \mathrm{~kg}}$

$\mathrm{Now, molality =\frac{\text { moles of } \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O_{6}}}{\text { mass of solvent }}=\frac{0.375}{182.5 \times 10^{-3}}$