250 g solution of D-glucose in water contain | vedclass

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(B) The molecular formula of glucose is $C_6H_{12}O_6$. The molar mass of glucose is $6 	imes 12 + 12 	imes 1 + 6 	imes 16 = 180 \ g/mol$.
The mas

Vedclass · Accepted Answer

$2.06$

Vedclass · Answer

(B) The molecular formula of glucose is $C_6H_{12}O_6$. The molar mass of glucose is $6 	imes 12 + 12 	imes 1 + 6 	imes 16 = 180 \ g/mol$.
The mass of carbon in $180 \ g$ of glucose is $6 	imes 12 = 72 \ g$.
Given that the solution contains $10.8 \ \%$ carbon by weight,the mass of carbon in $250 \ g$ of solution is $\frac{10.8}{100} 	imes 250 = 27 \ g$.
Since $72 \ g$ of carbon is present in $180 \ g$ of glucose,the mass of glucose containing $27 \ g$ of carbon is $\frac{180}{72} 	imes 27 = 67.5 \ g$.
The mass of the solvent (water) is $250 \ g - 67.5 \ g = 182.5 \ g = 0.1825 \ kg$.
The number of moles of glucose is $\frac{67.5 \ g}{180 \ g/mol} = 0.375 \ mol$.
Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg} = \frac{0.375 \ mol}{0.1825 \ kg} \approx 2.055 \ m \approx 2.06 \ m$.

$250 \ g$ solution of $D$-glucose in water contains $10.8 \ \%$ of carbon by weight. The molality of the solution is nearest to (Given: Atomic weights are $H = 1 \ u, C = 12 \ u, O = 16 \ u$)

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