Answer: To calculate the number of moles of NaOH withdrawn from the stock solution, we can use the formula:
moles = concentration (in M) x volume (in L)
On Day 1, Ray and Polly prepared a 1.35 M aqueous solution of NaOH. This means that there are 1.35 moles of NaOH in every liter of the solution.
On Day 2, they took 30.0 mL of this stock solution and diluted it with water in a 120.0-mL volumetric flask. The final volume of the diluted solution is 120.0 mL.
To calculate the concentration of the diluted solution, we can use the formula:
concentration (in M) = moles / volume (in L)
Since we know that the final volume is 120.0 mL, or 0.120 L, we need to calculate the number of moles in the diluted solution.
moles = concentration (in M) x volume (in L)
moles = 1.35 M x 30.0 mL / 1000 mL/L
moles = 0.0405 moles
Therefore, Ray and Polly withdrew 0.0405 moles of NaOH from the stock solution
To determine the final molar concentration of the Day 2 solution, we can use the equation:
M1V1 = M2V2
Substituting these values into the equation, we get:
(1.35 M)(30.0 mL) = M2(120.0 mL)
Solving for M2, we get:
M2 = (1.35 M)(30.0 mL)/(120.0 mL)
M2 = 0.3375 M. Therefore, the final molar concentration of the Day 2 solution is 0.3375 M.
6. MS10Q6: To calculate the volume of the stock solution required to make a specific volume of diluted solution, we can use the following formula:
V1 = (C2 x V2) / C1
Where:
V1 = Volume of stock solution required
C1 = Concentration of stock solution
V2 = Final volume of diluted solution
C2 = Desired concentration of diluted solution
Using the values given in the question, we can calculate the volume of the 1.900 M NaCl solution that Jesus must use in order to make 2.819 L of 0.224 M NaCl solution:
V1 = (0.224 M x 2.819 L) / 1.900 M
V1 = 0.333 L or 333 mL
Therefore, Jesus must use 333 mL or 0.333 L of the 1.900 M NaCl solution to make 2.819 L of 0.224 M NaCl solution.