A very dilute solution is prepared by dissolving 'x1' mole of solute in 'x2' mole of a solvent. The mole fraction of solute is approximately equal to
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Here's how you can approach such problems.
Explanation:
You know that your solution is 38% w/whydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.
To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.
Use the solution's density to determine what the mass of this sample would be
1.00L⋅1000mL1L⋅1.19 g1mL=1190 g
Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get
1190g solution⋅38 g HCl100g solution=452.2 g HCl
To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.
452.2g⋅1 mole36.46g=12.40 moles
Since the sample has a volume of 1.00-L, the molarity will be
C=nV=12.40 moles1.00 L=12.4 M
To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write
msol=mwater+macid
mwater=msol−macid=1190−452.2=737.8 g
The molality will thus be - do not forget to convert the mass of water to kilograms!
b=nmwater=12.40 moles737.8⋅10−3kg=16.8 molal
Since hydrochloric acid, HCl, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.
normality=Cfeq, where
C - the solution's molarity;
feq - the equivalence factor, in your case equal to 1.
normality=12.40 M1=12.4 N
Here's how you can approach such problems.
Explanation:
You know that your solution is 38% w/whydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.
To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.
Use the solution's density to determine what the mass of this sample would be
1.00L⋅1000mL1L⋅1.19 g1mL=1190 g
Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get
1190g solution⋅38 g HCl100g solution=452.2 g HCl
To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.
452.2g⋅1 mole36.46g=12.40 moles
Since the sample has a volume of 1.00-L, the molarity will be
C=nV=12.40 moles1.00 L=12.4 M
To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write
msol=mwater+macid
mwater=msol−macid=1190−452.2=737.8 g
The molality will thus be - do not forget to convert the mass of water to kilograms!
b=nmwater=12.40 moles737.8⋅10−3kg=16.8 molal
Since hydrochloric acid, HCl, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.
normality=Cfeq, where
C - the solution's molarity;
feq - the equivalence factor, in your case equal to 1.
normality=12.40 M1=12.4 N
Explanation:
You know that your solution is 38% w/whydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.
To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.
Use the solution's density to determine what the mass of this sample would be
1.00L⋅1000mL1L⋅1.19 g1mL=1190 g
Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get
1190g solution⋅38 g HCl100g solution=452.2 g HCl
To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.
452.2g⋅1 mole36.46g=12.40 moles
Since the sample has a volume of 1.00-L, the molarity will be
C=nV=12.40 moles1.00 L=12.4 M
To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write
msol=mwater+macid
mwater=msol−macid=1190−452.2=737.8 g
The molality will thus be - do not forget to convert the mass of water to kilograms!
b=nmwater=12.40 moles737.8⋅10−3kg=16.8 molal
Since hydrochloric acid, HCl, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.
normality=Cfeq, where
C - the solution's molarity;
feq - the equivalence factor, in your case equal to 1.
normality=12.40 M1=12.4 N
Here's how you can approach such problems.
Explanation:
You know that your solution is 38% w/whydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.
To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.
Use the solution's density to determine what the mass of this sample would be
1.00L⋅1000mL1L⋅1.19 g1mL=1190 g
Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get
1190g solution⋅38 g HCl100g solution=452.2 g HCl
To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.
452.2g⋅1 mole36.46g=12.40 moles
Since the sample has a volume of 1.00-L, the molarity will be
C=nV=12.40 moles1.00 L=12.4 M
To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write
msol=mwater+macid
mwater=msol−macid=1190−452.2=737.8 g
The molality will thus be - do not forget to convert the mass of water to kilograms!
b=nmwater=12.40 moles737.8⋅10−3kg=16.8 molal
Since hydrochloric acid, HCl, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.
normality=Cfeq, where
C - the solution's molarity;
feq - the equivalence factor, in your case equal to 1.
normality=12.40 M1=12.4 N
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