Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
Answer:
0.143 g of KCl.
Explanation:
Equation of the reaction:
AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)
Molar concentration = mass/volume
= 0.16 * 0.012
= 0.00192 mol AgNO3.
By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.
Number of moles of KCl = 0.00192 mol.
Molar mass of KCl = 39 + 35.5
= 74.5 g/mol
Mass = molar mass * number of moles
= 74.5 * 0.00192
= 0.143 g of KCl.
The answer would be 0.55 moles! good luck!