Calculate the mass of Glucose (C_6H_{12}O_6) | vedclass

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

Question

(B) The molar mass of Glucose $(C_6H_{12}O_6)$ is $(6 	imes 12) + (12 	imes 1) + (6 	imes 16) = 72 + 12 + 96 = 180 \ g/mol$. Molality $(m)$ is defi

Vedclass · Accepted Answer

$107.65$

Vedclass · Answer

(B) The molar mass of Glucose $(C_6H_{12}O_6)$ is $(6 	imes 12) + (12 	imes 1) + (6 	imes 16) = 72 + 12 + 96 = 180 \ g/mol$. Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent. Given $m = 0.25 \ mol/kg$. Let the mass of the solvent (water) be $W \ kg$. The mass of the solution is $W + 	ext{mass of glucose} = 2.5 \ kg$. Mass of glucose $= 	ext{moles} 	imes 	ext{molar mass} = (0.25 	imes W) 	imes 180 = 45W \ g = 0.045W \ kg$. Substituting this into the solution mass equation: $W + 0.045W = 2.5 \ kg$. $1.045W = 2.5 \implies W = 2.5 / 1.045 \approx 2.3923 \ kg$. Mass of glucose $= 0.045 	imes 2.3923 \ kg \approx 0.10765 \ kg = 107.65 \ g$.

Calculate the mass of Glucose $(C_6H_{12}O_6)$ required in making $2.5 \ kg$ of $0.25 \ molal$ aqueous solution. [Atomic wt : $H=1, O=16, C=12 \ amu$] (in $g$)

Similar Questions

If the density of a $1 \, m$ aqueous solution of $NaCl$ is $1.21 \, g/mL$,then the molarity of the solution is .......... (in $, M$)

The amount of ammonium sulphate required for the preparation of $1 \, M$ solution in $2 \, L$ is ............ $gm$.

Calculate the molarity of $NaOH$ in the solution prepared by dissolving $4 \,g$ of it in enough water to form $250 \,mL$ of the solution. (in $,M$)

$9.8 \ g$ of $H_2SO_4$ is present in $2 \ L$ of a solution. The molarity of the solution is $....... \ M$

With an increase in temperature,which of the following concentration terms changes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module