Question

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A. What volume of 2.50 M lead(II)nitrate solution contains 0.0500 mol of Pb2+? (

b. How many milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution? (

c. If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution?

Answer 1

what volume of 2.50 m lead ll nitrato

Answer 2

For A: The volume of solution is 0.02 L

For B: The volume of concentrated
`K_2Cr_2O_7` solution is 5 mL

For C: The molarity of diluted NaOH is 0.4 M

Step-by-step explanation:

• For A:

To calculate the volume for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

Molarity of lead (II) nitrate solution = 2.50 M

Moles of lead ions = 0.0500 mol

Putting values in above equation, we get:

`2.50M=\frac{0.0500mol}{\text{Volume of solution}}\\\\\text{Volume of solution}=(0.0500mol)/(2.50mol/L)=0.02L`

Hence, the volume of solution is 0.02 L

• For B:

To calculate the molarity of the diluted solution, we use the equation:

`M_1V_1=M_2V_2` ......(1)

where,

`M_1\text{ and }V_1` are the molarity and volume of the concentrated solution

`M_2\text{ and }V_2` are the molarity and volume of diluted solution

We are given:

`M_1=5.0M\\V_1=?mL\\M_2=0.01M\\V_2=250mL`

Putting values in equation 1, we get:

`5.0* V_1=0.10* 250\\\\V_1=5mL`

Hence, the volume of concentrated
`K_2Cr_2O_7` solution is 5 mL

• For C:

We are given:

`M_1=10.0M\\V_1=10.0mL\\M_2=?M\\V_2=250mL`

Putting values in equation 1, we get:

`10.0* 10.0=M_2* 250\\\\M_2=0.4M`

Hence, the molarity of diluted NaOH is 0.4 M