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3 years ago

Calculate the value of 0.1394÷0.004 showing all workings

Chemistry

1 answer:

Lesechka [4]3 years ago

3 0

Answer:

34.85

Explanation:

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A 12.0 wt% solution of CaCl2 (110.98 g/mol) has a density of 1.107 g/mL.

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Answer:

a) Mass of solution is 16,605 milli grams.

b) 57.29 grams is the mass of calcium chloride in 431.3 mL of a 12.0 wt% solution.

c) 1.1967 mol/L is the formal concentration of calcium chloride in 431.3 mL solution.

Explanation:

Mass by mass percentage of calcium chloride solution, wt% = 12.0%

Density of the solution = d = 1.107 g/mL

a) Volume of the solution , V= 15.0 mL

Mass of the solution = m

$Mass=density\times Volume$

$m=d\times V=1.107 g/ml\times 15.0 mL=16.605 g$

Mass of solution = 16.605 g = 16,605 mg (1g = 1000 mg)

b) Mass of the solution = m

Volume of the solution = V = 431.3 mL

$Mass=density\times Volume$

$m=d\times V=1.107 g/ml\times 431.3 mL=477.4491 g$

Mass of calcium chloride = x

w% = 12.0%

$\frac{x}{M}\times 100=12\%$

$x=12\times \frac{M}{100}=12\times \frac{477.4491 g}{100}=57.29 g$

57.29 grams is the mass of calcium chloride in 431.3 mL of a 12.0 wt% solution.

c) Moles of calcium chloride = $\frac{57.99 g}{111 g/mol}=0.5161 mol$

$Molarity(M)=\frac{moles(n)}{V(L)}$

V = Volume of the solution in Liters

n = 0.5161 mole

V = 431.3 mL = 0.4313 L

$M=\frac{0.5161 mol}{0.4313 L}=1.1967 mol/L$

1.1967 mol/L is the formal concentration of calcium chloride in 431.3 mL solution.

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Calculate the value of 0.1394÷0.004 showing all workings​

Calculate the value of 0.1394÷0.004 showing all workings