JEE MAIN - Chemistry (2022 - 27th July Morning Shift - No. 1)
$$250 \mathrm{~g}$$ 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, $$\mathrm{H}, 1 \,\mathrm{u} ; \mathrm{C}, 12 \,\mathrm{u} ; \mathrm{O}, 16 \,\mathrm{u}$$)
1.03
2.06
3.09
5.40
Explanation
Weight of D-glucose in water $=250 \mathrm{~g}$
$$ \begin{aligned} \therefore \text { Weight of carbon in D-glucose } &=\frac{250}{180} \times 72 \\\\ &=100 \mathrm{~g} \end{aligned} $$
$\%$ of carbon in the aqueous solution of glucose is $=10.8 \%$
$\therefore$ Weight of the solution is $=925.93$
$$ \begin{aligned} \therefore \text { Molality of D-glucose is } &=\frac{\frac{250}{180}}{(925.93-250)} \times 1000 \\\\ &=\frac{250}{180 \times 675.93} \times 1000 \\\\ &=2.06 \end{aligned} $$
$$ \begin{aligned} \therefore \text { Weight of carbon in D-glucose } &=\frac{250}{180} \times 72 \\\\ &=100 \mathrm{~g} \end{aligned} $$
$\%$ of carbon in the aqueous solution of glucose is $=10.8 \%$
$\therefore$ Weight of the solution is $=925.93$
$$ \begin{aligned} \therefore \text { Molality of D-glucose is } &=\frac{\frac{250}{180}}{(925.93-250)} \times 1000 \\\\ &=\frac{250}{180 \times 675.93} \times 1000 \\\\ &=2.06 \end{aligned} $$
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