An aqueous solution of Pb(NO3)2 is made by placing 275 g of solid Pb(NO3)2 into a volumetric flask and adding water to the 1.00 L mark (assume that 775g of water has been added to achieve this total solution volume). (Assume MW of Pb(NO3)2 = 331g/mole) A) What is the molarity (M) of this solution? B) What is the molality (m) of this solution? C) What is the mass % of Pb(NO3)2 in this solution? D) What is the mole fraction of Pb(NO3)2 present in this solution?

A) 0.831 M

B) 1.07 m

C) 26.2%

D) 0.0189

Explanation:

A) First we convert the mass of Pb(NO₃)₂ to moles:

Then we divide it by the total volume (1.00 L) to calculate the molarity:

B) We convert the grams of water to kilograms:

Then we divide the Pb(NO₃)₂ by the kilograms of water:

C) We divide the mass of Pb(NO₃)₂ by the total mass of the solution:

D) We calculate the moles of water:

Then we divide the Pb(NO₃)₂ moles by the total number of moles:

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.