Answer:
552 g of LiNO₃
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mass of LiNO₃ =?
Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mole of LiNO₃ =?
Molarity = mole /Volume
4 = mole of LiNO₃ / 2
Cross multiply
Mole of LiNO₃ = 4 × 2
Mole of LiNO₃ = 8 moles
Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:
Mole of LiNO₃ = 8 moles
Molar mass of LiNO₃ = 7 + 14 + (16×3)
= 7 + 14 + 48
= 69 g/mol
Mass of LiNO₃ =?
Mole = mass /Molar mass
8 = Molar mass of LiNO₃ /69
Cross multiply
Molar mass of LiNO₃ = 8 × 69
Molar mass of LiNO₃ = 552 g
Thus, 552 g of LiNO₃ is needed to prepare the solution.
Answer:
23.8
Explanation:
Formula
weight % = weight of solute/ weight of solution x 100
weight of solution = weight of salt + weight of water
weight of solution = 1.62 lb + 5.20 lb = 6.82 lb
weight % = 1.62 / 6,82 x 100
weight % = 0.238 x 100
weight % = 23.8
