Molarity problem using specific gravity?
I need an explanation for the following problem and answer. My book has an explanation I don't understand.
Problem: " "What is the approximate molarity of NaCl solution with a specific gravity of 1.006?"
Answer: 0.1M
The book's explanation says that 1L of water weighs 1kg so lL of solution weighs 1.006 kg. I thought specific gravity was in g/ml, so I assumed 1006g/L. The book then says if we assume the volume changes very little by the addition of NaCl then about .006kg or 6g of NaCl are in each liter of solution. Am I crazy or is the book wrong? I've spent over an hour on the net trying to find an explanation, so any info would be appreciated. Thanks.
Answer:
1006 g is the weight of one liter of solution
one liter of water weights 1000 g
1006 - 1000 = 6 grams of solute added
Molecular weight NaCl = 58.45 g/mol
6 / 58.45 = 0.1 moles NaCl
Molarity = moles solute / L solution = 0.1 / 1.0 = 0.1 M
You can assume that density is = to specific gravity
For water density = 1.0 g/ ml
D = M / V ; where D = density , M = mass of solution (g ) ,
V = volume of solution ( ml )
Since the volume of the solution is not stated in the problem we will assume that the volume of the solution is 1000 ml or 1 liter
Therefore the mass of water = 1000 grams
The mass of solution ; M = DV = ( 1.006 g/ml ) ( 1000 ml ) = 1006 grams
The mass of NaCl = mass of solution – mass 0f water = 1006 – 1000 = 6 grams
Molarity = moles of solute/ vol of solution ( L)
= 6/58.45/ 1liter = 0.10 M ans.
S.G. has no units. The S.G. of a substance indicates its Density compared to the Density of the same volume of water (or air if a gas).
Water density is 1,000g/L.
NaCl at a S.G. of 1.006 = 1,000g/L x 1.006 = 1006g/L.
= 6g NaCl/L of solution.
Molarity (M) = g/L
NaCl Mol.mass = 23 + 35.5 = 58.5g/mol.
Moles = Mass ÷ Mol.mass
Molarity = 6g/L ÷ 58.5g/mol. = 0.10
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Problem: " "What is the approximate molarity of NaCl solution with a specific gravity of 1.006?"
Answer: 0.1M
The book's explanation says that 1L of water weighs 1kg so lL of solution weighs 1.006 kg. I thought specific gravity was in g/ml, so I assumed 1006g/L. The book then says if we assume the volume changes very little by the addition of NaCl then about .006kg or 6g of NaCl are in each liter of solution. Am I crazy or is the book wrong? I've spent over an hour on the net trying to find an explanation, so any info would be appreciated. Thanks.
Answer:
1006 g is the weight of one liter of solution
one liter of water weights 1000 g
1006 - 1000 = 6 grams of solute added
Molecular weight NaCl = 58.45 g/mol
6 / 58.45 = 0.1 moles NaCl
Molarity = moles solute / L solution = 0.1 / 1.0 = 0.1 M
You can assume that density is = to specific gravity
For water density = 1.0 g/ ml
D = M / V ; where D = density , M = mass of solution (g ) ,
V = volume of solution ( ml )
Since the volume of the solution is not stated in the problem we will assume that the volume of the solution is 1000 ml or 1 liter
Therefore the mass of water = 1000 grams
The mass of solution ; M = DV = ( 1.006 g/ml ) ( 1000 ml ) = 1006 grams
The mass of NaCl = mass of solution – mass 0f water = 1006 – 1000 = 6 grams
Molarity = moles of solute/ vol of solution ( L)
= 6/58.45/ 1liter = 0.10 M ans.
S.G. has no units. The S.G. of a substance indicates its Density compared to the Density of the same volume of water (or air if a gas).
Water density is 1,000g/L.
NaCl at a S.G. of 1.006 = 1,000g/L x 1.006 = 1006g/L.
= 6g NaCl/L of solution.
Molarity (M) = g/L
NaCl Mol.mass = 23 + 35.5 = 58.5g/mol.
Moles = Mass ÷ Mol.mass
Molarity = 6g/L ÷ 58.5g/mol. = 0.10
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