2. Suppose a chemistry student made 2000. mL of 6.00 M HCl. He accidently spilled half of this solution on the floor. What is the Molarity of HCL left in the beaker? Explain the answer...
- for this problem, would I use the eq'N, M1V1=M2V2.
3. What is the molarity of a sol'n that contains 28.5 g of KOH in 2.00 L of sol'n?
5. If 500. ml of 2.0 M H2SO4 is diluted with water to a volume of 1.00 L, what is the molarity of the new solution.
2. The lost solution does not change the molarity of what is left in the beaker.
3. M=(28.5/40)/2
4. you are diluting it by a factor of 2. New molarity=1M
2. I still don't get this or is it the same equation.
3. Answer: 0.254 M KOH
5. Am I still using the equation?
Is #5's answer: 1.0 M H2SO4?
what about #2?
2. To find the molarity of HCl left in the beaker after spilling half of the solution, you can use the equation M1V1 = M2V2.
M1 represents the initial molarity of the solution, which is 6.00 M. V1 represents the initial volume of the solution, which is 2000 mL.
Since the student spilled half of the solution, the remaining volume in the beaker is 2000 mL / 2 = 1000 mL.
To use the equation, you need to convert the volumes to liters.
V1 = 2000 mL = 2.00 L (divide by 1000)
V2 = 1000 mL = 1.00 L (divide by 1000)
Now let's plug in the values into the equation:
M1V1 = M2V2
6.00 M * 2.00 L = M2 * 1.00 L
12.00 = M2 * 1.00
Divide both sides by 1.00 to solve for M2:
M2 = 12.00 / 1.00
M2 = 12.00 M
Therefore, the molarity of HCl left in the beaker is 12.00 M.
3. To find the molarity of a solution, you need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, you need to calculate the number of moles of KOH using the given mass and the molar mass of KOH.
Molar mass of KOH = 39.10 g/mol (K) + 16.00 g/mol (O) + 1.01 g/mol (H) = 56.11 g/mol
moles of KOH = mass of KOH / molar mass of KOH
moles of KOH = 28.5 g / 56.11 g/mol
Now you can calculate the molarity using the given volume of solution:
Molarity = moles of KOH / volume of solution
Molarity = moles of KOH / 2.00 L
Plug in the values:
Molarity = 28.5 g / 56.11 g/mol / 2.00 L
Solve this expression to find the molarity.
5. To find the molarity of the diluted solution, you can use the formula:
M1V1 = M2V2
M1 represents the initial molarity of the solution, which is 2.0 M. V1 represents the initial volume of the solution, which is 500 mL. V2 represents the final volume of the solution after dilution, which is 1.00 L.
First, convert the volumes to liters:
V1 = 500 mL = 0.500 L (divide by 1000)
V2 = 1.00 L (no conversion needed)
Now let's plug in the values into the equation:
M1V1 = M2V2
2.0 M * 0.500 L = M2 * 1.00 L
1.00 = M2 * 1.00
Solve for M2:
M2 = 1.00 / 1.00
M2 = 1.00 M
Therefore, the molarity of the new solution after dilution is 1.00 M.