Calculate the percent by mass of the solute in each of the following aqueous solutions: (a) 5.50 g of NaBr in 78.2 g of solution, (b) 31.0 g of KCl in 152 g of water, (c) 4.5 g of toluene in 29 g of benzene.
Answers
Answer:
For a: The mass percent of NaBr is 7.03 %
For b: The mass percent of KCl is 16.94 %
For c: The mass percent of toluene is 13.43 %
Explanation:
To calculate the mass percentage of solute in solution, we use the equation:
.......(1)
- For a:
We are given:
Mass of NaBr (Solute) = 5.50 g
Mass of solution = 78.2 g
Putting values in equation 1, we get:
Hence, the mass percent of NaBr is 7.03 %
- For b:
We are given:
Mass of KCl (Solute) = 31.0 g
Mass of water (solvent) = 152 g
Mass of solution = (31.0 + 152) g = 183 g
Putting values in equation 1, we get:
Hence, the mass percent of KCl is 16.94 %
- For c:
We are given:
Mass of toluene (Solute) = 4.5 g
Mass of benzene (solvent) = 29 g
Mass of solution = (4.5 + 29) g = 33.5 g
Putting values in equation 1, we get:
Hence, the mass percent of toluene is 13.43 %
Final answer:
To calculate the percent by mass of the solute in each aqueous solution, divide the mass of the solute by the mass of the solution and multiply by 100%
Explanation:
To calculate the percent by mass of the solute in each aqueous solution, you'll need to use the formula:
Percent by mass = (mass of solute/mass of solution) x 100%
For example, in solution (a) with 5.50 g of NaBr in 78.2 g of solution, the mass of the solute is 5.50 g and the mass of the solution is 78.2 g. Plugging these values into the formula gives:
Percent by mass = (5.50 g / 78.2 g) x 100% = 7.03%
Similarly, you can calculate the percent by mass for solutions (b) and (c) using the same formula.
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Sulfur usually is on -2, +4 or +6 level of oxidation (but it can be also on 5, 3, 2, 1, -1 ).
Out of given choices (B) is the most likely one.
Answers
Answer: B. Ca(OH)2
Explanation:just took the quiz/quick check
This reaction is best described as
(1) addition involving a saturated hydrocarbon
(2) addition involving an unsaturated hydrocarbon
(3) substitution involving a saturated hydrocarbon
(4) substitution involving an unsaturated hydrocarbon
Answers
This reaction is best described as substitution involving a saturated hydrocarbon. Therefore, the correct option is option C.
When one atom or group of atoms in a molecule is replaced by another atom or group of atoms, the chemical reaction is known as a substitution reaction. Usually, a reactant molecule and a reagent molecule, which supplies the replacing atom or group, engage in this reaction. Nucleophilic substitution and electrophilic substitution are the two primary categories of substitution processes. A nucleophile (a substance rich in electrons) can replace another atom or group in a molecule in nucleophilic substitution processes. A nucleophile combines with an electrophile (an electron-deficient species) in organic chemistry regularly to produce this kind of reaction, which results in the formation of a new molecule.
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involving a saturated hydrocarbon. You
have substitued a chlorine atom for one of the hydrogen atoms, and the hydrocarbon you start with (ethane) is fully saturated.
Answers
Answers
Now to find number of particles we take number of moles x avogadros number.
1.5 x (6.02x10^23) = 9.03x10^23 molecules
is required?
a. 7.00 mL
b. 8.40 mL
c. 17.1 mL
d. 58.3 mL
Answers
Answer:
Explanation:
To determine the volume of the stock solution required to prepare 3.50 L of 0.200 M hydro chloric acid, we can use the formula:
M1V1 = M2V2
where:
M1 = concentration of the stock solution
V1 = volume of the stock solution
M2 = desired concentration of the diluted solution
V2 = desired volume of the diluted solution
Let's substitute the given values into the formula:
M1 = 12.0 M
V1 = ?
M2 = 0.200 M
V2 = 3.50 L
Now we can solve for V1:
12.0 M x V1 = 0.200 M x 3.50 L
V1 = (0.200 M x 3.50 L) / 12.0 M
V1 = 0.0583 L
To convert the volume from liters to milliliters, we multiply by 1000:
V1 = 0.0583 L x 1000 mL/L
V1 = 58.3 mL
Therefore, the volume of the stock solution required is 58.3 mL.
So, the correct answer is d. 58.3 mL.
To determine the volume of the stock solution required, we can use the formula:
Molarity1 x Volume1 = Molarity2 x Volume2
Where Molarity1 and Volume1 represent the initial concentration and volume, and Molarity2 and Volume2 represent the final concentration and volume.
Given:
Molarity1 = 12.0 M
Volume1 = ?
Molarity2 = 0.200 M
Volume2 = 3.50 L
Plugging in the values into the formula, we have:
12.0 M x Volume1 = 0.200 M x 3.50 L
Simplifying the equation, we can solve for Volume1:
Volume1 = (0.200 M x 3.50 L) / 12.0 M
Volume1 ≈ 0.0583 L
To convert this to milliliters, we multiply by 1000:
Volume1 ≈ 58.3 mL
Therefore, the volume of the stock solution required is approximately 58.3 mL.
The closest answer option is d. 58.3 mL.
I hope this explanation helps! Let me know if you have any further questions.