(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?

For A: The volume of solution is 0.02 L

For B: The volume of concentrated solution is 5 mL

For C: The molarity of diluted NaOH is 0.4 M

Explanation:

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

Molarity of lead (II) nitrate solution = 2.50 M

Moles of lead ions = 0.0500 mol

Putting values in above equation, we get:

Hence, the volume of solution is 0.02 L

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

      ......(1)

where,

are the molarity and volume of the concentrated solution

are the molarity and volume of diluted solution

We are given:

Putting values in equation 1, we get:

Hence, the volume of concentrated solution is 5 mL

We are given:

Putting values in equation 1, we get:

Hence, the molarity of diluted NaOH is 0.4 M

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).

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