Question

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?

Answer 1

Answer:

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:

• 275 g ÷ 331 g/mol = 0.831 mol Pb(NO₃)₂

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

• 0.831 mol / 1.00 L = 0.831 M

B) We convert the grams of water to kilograms:

• 775 g / 1000 = 0.775 kg

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

• 0.831 mol / 0.775 kg = 1.07 m

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

• $\frac{gPb(NO_{3})_{2}}{gPb(NO_{3})_{2}+gH_{2}O}$ *100% $=\frac{275}{275+775}$ * 100% = 26.2%

D) We calculate the moles of water:

• 775 g H₂O ÷ 18g/mol = 43.1 mol H₂O

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

• $\frac{0.831}{0.831+43.1}$ = 0.0189