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3 years ago

Calcular la fraccion molar de NaClO en una solucion blanqueadora comercial que comtiene 3,62% en masa de NaClO en agua

1 answer:

muminat3 years ago

7 0

Calculate the molar fraction of NaClO in a commercial bleach solution that contains 3.62% by mass of NaClO in water

<h3>Further explanation</h3>

convert mass to mol

mol NaClO (MW=74.5 g/mol) :

$\tt \dfrac{3.62}{74.5}=0.0486$

mol H₂O (MW= 18 g/mol) :

$\tt \dfrac{96.38}{18}=5.354$

Total mol :

0.0486 + 5.354 = 5.4026

mol fraction of NaClO :

$\tt X_{NaClO}=\dfrac{0.0486}{5.4026}=0.009=9.10^{-3}$

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3 years ago

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4 years ago

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2. A solution is made by adding 1.23 mol of KCl to 1000.0 g of water. Assume that the

ladessa [460]

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a. 74.55 g/mol.

b. 91.70 g.

c. 8.40%.

d. 1.23 mol/L.

Explanation:

a. Calculate the formula weight of KCl.

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atomic weight of K = 39.098 g/mol, atomic weight of Cl = 35.45 g/mol.

∴ Formula weight of KCl = atomic weight of K + atomic weight of Cl = 39.098 g/mol + 35.45 g/mol = 74.548 g/mol ≅ 74.55 g/mol.

b. Calculate the mass of KCl in grams.

we can use the relation:

no. of moles (n) = mass/molar mass.

∴ mass of KCl = n*molar mass = (1.23 mol)*(74.55 g/mol) = 91.69 g ≅ 91.70 g.

c. Calculate the percent by mass of KCl in this solution.

The mass % of KCl = (mass of KCl/mass of the solution) * 100.

mass of KCl = 91.70 g,

mass of the solution = 1000.0 g of water + 91.70 g of KCl = 1091.70 g.

∴ The mass % of KCl = (91.70 g/1091.70 g)*100 = 8.399% ≅ 8.40%.

d. Calculate the molarity of the solution.

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of KCl)/(Volume of the solution (L))

no. of moles of KCl = 1.23 mol,

Volume of the solution = mass of water / density of water = (1000.0 g)/(1.00 g/mL) = 1000.0 mL = 1.0 L.

M = (1.23 mol)/(1.0 L) = 1.23 mol/L.

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4 years ago