What volume (L) of an aqueous solution that is 12.0 % by mass arsenic acid (H 3
ID: 824217 • Letter: W
Question
What volume (L) of an aqueous solution that is 12.0 % by mass arsenic acid (H3AsO4) and has a density of 1.08 g/mL must be taken to obtain 36.9 g of arsenic acid (H3AsO4).
A 5760 mL aqueous solution of ammonia (NH3) has a density of 0.910 g/mL and contains 1260 g of ammonia (NH3). Calculate the fraction by mass of ammonia (NH3) in the solution.
An aqueous solution that is 25.0 % by mass glycerol (C3H8O3) has a density of 1.06 g/mL. Calculate the mass (g) of glycerol (C3H8O3) in precisely 1 L of solution.
Calculate the mass (g) of arsenic acid (H3AsO4) present in 0.329 L of an aqueous solution that is 12.0% by mass arsenic acid (H3AsO4) and has a density of 1.08 g/mL.
A 6380 mL aqueous solution contains 3890 g of formic acid (HCO2H) and is 54.0 % by mass formic acid (HCO2H). Calculate the density (g/mL) of the solution of formic acid (HCO2H).
A 247 mL aqueous solution contains 32.0 g of arsenic acid (H3AsO4) and is 12.0 % by mass arsenic acid (H3AsO4). Calculate the density (g/mL) of the solution of arsenic acid (H3AsO4).
Molar Mass (g/mol)
Molar Mass (g/mol) H3AsO4 141.94Explanation / Answer
a. Let the mass of solution be x g.
Mass of arsenic, 36.9=12.0x/100
x=307.5g
Volume= mass/density
=307.5/1.08=284.72 ml
b. Mass of solution= volume*density
=5760*0.910=5241.6 g
Mass fraction of ammonia=mass of NH3/total mass
=1260/5241.6
=0.240 or 24.0%
c. Mass of 1L solution=1000*1.06
=1060g
Mass of glycerol=25% of 1060
=265g
d. Mass of 0.329L solution=329*1.08
=355.32g
Mass of arsenic acid=12% of 355.32
=42.64g
e. Let mass of solution be x.
54% of x=3890g
x=7203.7g
Density=mass/volume
=7203.7/6380
=1.129 g/ml
f. Let mass of solution be x.
12% of x=32g
x=266.67g
Density=mass/volume
=266.67/247
=1.079 g/ml
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.