A0.450 g sample of impure caco3(s) is dissolved in 50.0 ml of 0.150 m hcl ( aq ) . the equation for the reaction is caco3 ( s ) + 2 hcl ( aq ) ⟶ cacl2 ( aq ) + h2o ( l ) + co2 ( g ) the excess hcl ( aq ) is titrated by 5.35 ml of 0.125 m naoh ( aq ) . calculate the mass percentage of caco3(s) in the sample.
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Ответ:
The mass percentage of CaCO3(s) in the sample is 83.4%
Explanation:
CaCO₃ + 2HCl → CaCl₂ + H2O + CO₂
Let's find out the moles of each reactant:
[HCl] = 0.150 mol/L
Molarity . volume = moles
0.150 m/L . 0.050L = 7.5x10⁻³ moles HCl
Molar mass of CaCO₃ =100.08 g
Moles of CaCO₃ = Mass / Molar mass
0.450 g/100.08 g = 4.49x10⁻³ moles of salt
Ratio is 1:2
If 2 moles of HCl react with 1 mol of salt
7.5x10⁻³ moles of HCl react with 7.5x10⁻³ / 2 = 3.75x10⁻³ moles
We must find out the mass
Moles of CaCO₃ . Molar mass CO₃ = 0.3753 g
So now, the mass percentage of salt will be
0.450g ___ 100%
0.3753 g (0.3753 .100 )/0.450 =83.4 %
Ответ:
Molality → 0.188 m
Mole fraction of fructose → 0.00337
Mass percent of fructose in solution → 3.29 %
Molarity → 0.183 M
Explanation:
Solute → 34 g of fructose
Solvent → 1000 g of water
Solution → 1000 g of water + 34 g of fructose = 1034 g of solution.
We take account density to calculate, the solution's density
1.0078 g/mL = 1034 g / mL
1034 g / 1.0078 g/mL = 1026 mL
Molal concentration → moles of solute in 1kg of solvent
Moles of fructose = mass of fructose / molar mass
34 g/ 180g/mol = 0.188 mol
0.188 mol/1kg = 0.188 m
Mole fraction of fructose = Moles of fructose / Total moles
We determine the moles of water
Moles of water = 1000 g / 18 g = 55.5 mol
Total moles = moles of fructose + moles of water
0.188 mol + 55.5 mol = 55.743 mol
0.188 mol / 55.743 mol = 0.00337
Mass percent = mass of fructose in 100 g of solution
(Mass of fructose / Total mass ) . 100 = (34 g /1034 g) . 100 = 3.29 %
Molarity = Moles of solute in 1L of solution
We can also say mmol of solute in 1 mL of solution
0.188 mol of fructose = 188 mmol of fructose
Molarity = 188 mmol / 1026 mL of solution = 0.183 M