It is impossible to calculate the molarity of a solution prepared by dissolving 25 g sugar in 500 g of water . These figures are normally used when dealing with molaLity of solutions . Not molarity.
I will calculate molality :
Molality = moles of solute dissolved in 1.0 kg water .
You have 25 g sugar dissolved in 500 g water
This is the same as 50 g sugar dissolved in 1000 g water or 1.0 kg water .
Molar mass sugar = 342 g/mol
Mol sugar in 50 g = 50 g / 342 g/mol = 0.146 mol dissolved in 1.0 kg water
It is impossible to calculate the molarity of a solution prepared by dissolving 25 g sugar in 500 g of water . These figures are normally used when dealing with molaLity of solutions . Not molarity.
I will calculate molality :
Molality = moles of solute dissolved in 1.0 kg water .
You have 25 g sugar dissolved in 500 g water
This is the same as 50 g sugar dissolved in 1000 g water or 1.0 kg water .
Molar mass sugar = 342 g/mol
Mol sugar in 50 g = 50 g / 342 g/mol = 0.146 mol dissolved in 1.0 kg water
There are many different types of sugars so assuming you meant glucose whose chemical formula is C6H12O6
molarity= moles of solute/volume of solution
moles= mass/molar mass
mass of C6H12O6=25g
molar mass of C6H12O6=180g/mol
so
moles of C6H12O6=25/180=0.13888mol
H2O is the solvent whose volume is found by using the formula
density= mass/volume
so
volume= mass/density
knowing that the density of H2O= 1g/ml
while the given mass of H2O= 500g
plugging these values into the formula
volume of H2O=500/1=500ml
converting ml to liters
(1000ml=1L)
(500ml)*(1L)/(1000ml)=500/1000 =0.5L
the molarity of the sugar solution=0.13888/0.5=0.28M
the mole fraction of C6H12O6 in the solution is found by using the formula
X(C6H12O6)= moles of C6H12O6/moles of C6H12O6 + moles of H2O
where
X represents the mole fraction of a substance,
X(C6H12O6)= unknown
moles of C6H12O6=0.28mol
the moles of H2O= given mass of H2O/molar mass of H2O
moles of H2O=500/18=28moles
so
X(C6H12O6)=0.28/0.28+28
X(C6H12O6)=0.01
the mole fraction of H2O is found by knowing that the sum of all the mole fractions in a solution must =1
so
the mole fraction of H2O is
X(H2O)= 1–0.01=0.99
There are many different types of sugars so assuming you meant glucose whose chemical formula is C6H12O6
molarity= moles of solute/volume of solution
moles= mass/molar mass
mass of C6H12O6=25g
molar mass of C6H12O6=180g/mol
so
moles of C6H12O6=25/180=0.13888mol
H2O is the solvent whose volume is found by using the formula
density= mass/volume
so
volume= mass/density
knowing that the density of H2O= 1g/ml
while the given mass of H2O= 500g
plugging these values into the formula
volume of H2O=500/1=500ml
converting ml to liters
(1000ml=1L)
(500ml)*(1L)/(1000ml)=500/1000 =0.5L
the molarity of the sugar solution=0.13888/0.5=0.28M
the mole fraction of C6H12O6 in the solution is found by using the formula
X(C6H12O6)= moles of C6H12O6/moles of C6H12O6 + moles of H2O
where
X represents the mole fraction of a substance,
X(C6H12O6)= unknown
moles of C6H12O6=0.28mol
the moles of H2O= given mass of H2O/molar mass of H2O
moles of H2O=500/18=28moles
so
X(C6H12O6)=0.28/0.28+28
X(C6H12O6)=0.01
the mole fraction of H2O is found by knowing that the sum of all the mole fractions in a solution must =1
so
the mole fraction of H2O is
X(H2O)= 1–0.01=0.99
More
VOTE
It is impossible to calculate the molarity of a solution prepared by dissolving 25 g sugar in 500 g of water . These figures are normally used when dealing with molaLity of solutions . Not molarity.
I will calculate molality :
Molality = moles of solute dissolved in 1.0 kg water .
You have 25 g sugar dissolved in 500 g water
This is the same as 50 g sugar dissolved in 1000 g water or 1.0 kg water .
Molar mass sugar = 342 g/mol
Mol sugar in 50 g = 50 g / 342 g/mol = 0.146 mol dissolved in 1.0 kg water
Molality = 0.146 m .
Mole fraction
Moles of sugar = 0.146 mol
Moles of water = 1000 g / 18 g/mol = 55.5 mol H2O
Total moles = 55.646
Mol fraction sugar = 0.146 /55.646= 0.0026
Mol fraction H2O = 55.5 / 55.646 = 0.9974
Mol fraction has no units
It is impossible to calculate the molarity of a solution prepared by dissolving 25 g sugar in 500 g of water . These figures are normally used when dealing with molaLity of solutions . Not molarity.
I will calculate molality :
Molality = moles of solute dissolved in 1.0 kg water .
You have 25 g sugar dissolved in 500 g water
This is the same as 50 g sugar dissolved in 1000 g water or 1.0 kg water .
Molar mass sugar = 342 g/mol
Mol sugar in 50 g = 50 g / 342 g/mol = 0.146 mol dissolved in 1.0 kg water
Molality = 0.146 m .
Mole fraction
Moles of sugar = 0.146 mol
Moles of water = 1000 g / 18 g/mol = 55.5 mol H2O
Total moles = 55.646
Mol fraction sugar = 0.146 /55.646= 0.0026
Mol fraction H2O = 55.5 / 55.646 = 0.9974
Mol fraction has no units
More
VOTE