25.00 g LiCl is dissolved in 85.00 g water to give 96.5 mL of solution. Calculat

Question

25.00 g LiCl is dissolved in 85.00 g water to give 96.5 mL of solution. Calculate the density of the solution. Calculate the % LiCl (by mass) in the solution. Calculate the molarity of the solution. 100.0 mL of Pb(NO_3)_2 solution is made by dissolving 47.41 g of Pb(NO_3)2(s) in 92.3 mL of water. The density of water is 0.9975 g/mL. Calculate the density of the solution. Calculate the % lead (Pb^2+) in the solution. Calculate the molarity of the solution. 50.00 g La(NO_3)_3 6H_2 O is dissolved in 50.00 g H_2 O to produce a solution with density 1.410 g/mL. Calculate the molarity of the solution. Calculate the % La(NO_3)_3 (not % La(NO_3)_3 6H_2 O) in the solution. An 80.00% sulfuric acid (aqueous) solution has density 1.7272 g/mL. Calculate the molarity of the solution. Calculate the volume of this solution that must be added to water to make 250.0 mL|of 3.00 M H_2 SO_4(aq). A 1.20 M Na_2 SO_4 (aqueous) solution has density 1.140 g/mL. Calculate the % Na_2 SO_4

Explanation / Answer

1)
a)
mass of solution = 25.00 g + 85.00 g = 110.0 g
volume of solution = 96.5 mL
density = mass/volume
= 110.0 g/ 96.5 mL
= 1.14 g/mL
Answer: 1.14 g/mL

b)
%LiCl = mass of LiCl * 100 / mass of solution
= 25.00*100/110.0
= 22.73 %

c)
number of moles of LiCl = mass of LiCl / molar mass of LiCl
= 25.00/42.4
=0.5896 mol

volume = 96.5 mL = 0.0965 L

M = number of mol of solute / volume of solution
= 0.5896 / 0.0965
= 6.11 M

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