A solution of cough syrup contains 5.00% active ingredient by volume. If the total volume of the bottle is 14.0 mL , how many mi
2 answers:
Answer : The volume of active ingredient present in the bottle are 0.7 mL
Explanation :
As we are give that:
5.00 % by volume means that 5.00 mL of active ingredient present in 100 mL of solution.
As, 100 mL of solution contains 5.00 mL of active ingredients
So, 14.0 mL of solution contains of active ingredients
Therefore, the volume of active ingredient present in the bottle are 0.7 mL
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Answer:
a. The maximum volume of 0.143 M HCl required is 154.4 mL.
b. The maximum volume of 0.143 M HCl required is 135.7 mL.
Explanation:
a.
Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)
Moles of aluminum hydroxide =
According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :
of HCl.
Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)
Moles of magnesium hydroxide =
According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :
of HCl.
Total moles of HCl required to neutralize both :
0.01346 mol + 0.008621 mol = 0.02208 mol
Molarity of the HCL solution = 0.143 M
Volume of the solution = V
1 L = 1000 mL
0.1544 L = 154.4 mL
The maximum volume of 0.143 M HCl required is 154.4 mL.
b.
Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)
Moles of calcium carbonate =
According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :
of HCl.
Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol
Molarity of the HCL solution = 0.143 M
Volume of the solution = V
1 L = 1000 mL
0.1357 L = 135.7 mL
The maximum volume of 0.143 M HCl required is 135.7 mL.