A solution contains 30 g of glucose , 20 g of salt in 500 ml of water. Calculate the mass percent of glucose and salt
Answers
The problem gives you all the information you need in order to solve for the molality and mole fraction of the solution. In order to determine its molarity, you're going to need the solution's volume.
To get the volume, you have to know what the density of the solution is. Determine the percent concentration by mass of the solution first
%w/w=msolutemsolution⋅100
In your case, the mass of the solution will be
msolution=mglucose+mwater
msolution=20+150=170 g
This means that you get
%w/w=20g170g⋅100=11.8%
The density of this solution will thus be
http://us.mt.com/us/en/home/supportive_content/application_editorials/D_Glucose_de_e.html
ρ=1.045 g/mL
Use glucose's molar mass to determine how many moles you have
20g⋅1 mole glucose180.16g=0.111 moles glucose
The solution's volume will be
170g⋅1 mL1.045g=162.7 mL
This means that its molarity is - do not forget to convert the volume to liters!
C=nV=0.111 moles162.7⋅10−3L=0.68 M
A solution's molality is defined as the number of moles of solute divided by the mass of the solvent - in kilograms! This means that you have
b=nmwater=0.111 moles150⋅10−3kg=0.74 molal
To get the mole fraction of sucrose, you need to know how many moles of water you have present. Once again, use water's molar mass
150g⋅1 mole water18.02g=8.24 moles water
The total number of moles the solution contains is
ntotal=nglucose+nwater
ntotal=0.111+8.24=8.351 moles
This means that the mole fraction of sucrose, which is defined as the number of moles of sucrose divided by the total number of moles in the solution, will be
χsucrose=nsucrosentotal=0.111moles8.351moles=0.013
Answer:
Explanation:
The problem gives you all the information you need in order to solve for the molality and mole fraction of the solution. In order to determine its molarity, you're going to need the solution's volume.
To get the volume, you have to know what the density of the solution is. Determine the percent concentration by mass of the solution first
%w/w=msolutemsolution⋅100
In your case, the mass of the solution will be
msolution=mglucose+mwater
msolution=20+150=170 g
This means that you get
%w/w=20g170g⋅100=11.8%
The density of this solution will thus
ρ=1.045 g/mL
Use glucose's molar mass to determine how many moles you have
20g⋅1 mole glucose180.16g=0.111 moles glucose
The solution's volume will be
170g⋅1 mL1.045g=162.7 mL
This means that its molarity is - do not forget to convert the volume to liters!
C=nV=0.111 moles162.7⋅10−3L=0.68 M
A solution's molality is defined as the number of moles of solute divided by the mass of the solvent - in kilograms! This means that you have
b=nmwater=0.111 moles150⋅10−3kg=0.74 molal
To get the mole fraction of sucrose, you need to know how many moles of water you have present. Once again, use water's molar mass
150g⋅1 mole water18.02g=8.24 moles water
The total number of moles the solution contains is
ntotal=nglucose+nwater
ntotal=0.111+8.24=8.351 moles
This means that the mole fraction of sucrose, which is defined as the number of moles of sucrose divided by the total number of moles in the solution, will be
χsucrose=nsucrosentotal=0.111moles8.351moles=0.