If 52.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.248 g of precipitate, wh
1 answer:
Explanation:
The volume of given lead nitrate solution is:
52.5 mL.
The amount of lead iodide formed is ---0.248 g.
To get the molarity of lead (II) ion follow the below-shown procedure:
The number of moles of lead iodide formed is:
0.000537 mole of lead iodide contains --- 0.000537 moles of lead (II) ion.
Thus, the number of moles are there, volume is there, and to get the molarity of lead (II) ion use the formula:
Molarity of lead iodide is --- 0.0102 M.
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Answer:
a) 0,2% w/v
b) r=500
c) 0,0182 M
d) 0,0145 m
e) 0,0137 equivalents
Explanation:
a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:
%w/v= ×100 = 0,2%w/v
b) Ratio strength is a way to express concentration. For w/v is in 1g of solute <em>r</em> mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.
c) Molarity is moles of solute per liter of solution, thus:
1,000 g of CaCl₂ × = 9,09×10⁻³ moles of CaCl₂
500 mL of solution × = 0,500 L of solution
M = = 0,0182 M
d) Molality is moles of solute per kg of solution.
Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:
kg of solution = 0,500 L of solution × =<em> </em><em>0,625 kg of solution</em>
m = = 0,0145 m
e) In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:
1,5 L × ×
= 0,0137 equivalents
I hope it helps!