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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $....$.

Similar Questions

$3.0 \ g$ ethanoic acid in $50 \ g$ benzene has . . . . . . molality. (Atomic weights: $H=1, C=12, O=16$).

The number of moles of solute per $kg$ of a solvent is called its

The density of a $1.0 \ M$ aqueous solution of glucose is $1.1 \ g/mL$. What is the molality of the solution in $m$ (in $m$)?

$A$ litre of seawater (which weighs $1030 \ g$) contains $6 \times 10^{-3} \ g$ of dissolved oxygen. The concentration of dissolved oxygen in $ppm$ is:

Molarity of a solution prepared by dissolving $75.5 \ g$ of pure $KOH$ in $540 \ mL$ solution is .......... $M$

Vedclass Test Series

Exam Paper Generator

Online Exam Module