The mole fraction of glucose (C_{6}H_{12}O_{6 | vedclass

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Question

(B) Given,mole fraction of glucose $X_{C_{6}H_{12}O_{6}} = 0.1$. Since it is a binary solution,the mole fraction of water $X_{H_{2}O} = 1 - 0.1 = 0.9$

Vedclass · Accepted Answer

$47$

Vedclass · Answer

(B) Given,mole fraction of glucose $X_{C_{6}H_{12}O_{6}} = 0.1$. Since it is a binary solution,the mole fraction of water $X_{H_{2}O} = 1 - 0.1 = 0.9$. Let the total number of moles be $1 \ mol$. Then,moles of glucose $n_{glucose} = 0.1 \ mol$ and moles of water $n_{water} = 0.9 \ mol$. Mass of glucose $= 0.1 \ mol 	imes 180 \ g/mol = 18 \ g$. Mass of water $= 0.9 \ mol 	imes 18 \ g/mol = 16.2 \ g$. Total mass of the solution $= 18 \ g + 16.2 \ g = 34.2 \ g$. Mass percentage of water $= (	ext{Mass of water} / 	ext{Total mass of solution}) 	imes 100 = (16.2 / 34.2) 	imes 100 \approx 47.36\%$. Rounding to the nearest integer,we get $47$.

The mole fraction of glucose $(C_{6}H_{12}O_{6})$ in an aqueous binary solution is $0.1$. The mass percentage of water in it,to the nearest integer,is $....$.

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