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nataly862011 [7]
3 years ago
8

(a) Calculate the mass of CaCl2·6H2O needed to prepare 0.125 m CaCl2(aq) by using 500. g of water. (b) What mass of NiSO4·6H2O m

ust be dissolved in 500. g of water to produce 0.22 m NiSO4(aq)?
Chemistry
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

(a) 13.7 g.

(b) 28.91 g.

Explanation:

  • molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.

∴ m = (no. of moles of solute)/(mass of water (kg))

<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>

<em />

<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>

∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).

m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).

∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.

<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>

∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).

m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.

∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).

∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.

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A student calculated the molarity of a solution prepared by dissolving 0.730 mol of table sugar (sucrose, C12H22O11) in 1.8x10^3
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Answer: C= 0.406 M

Explanation:

Solution.

ν

=

0.730

m

o

l

;

ν=0.730mol;

V

=

1.8

⋅

1

0

3

m

L

=

1.8

L

;

V=1.8⋅10

3 mL=1.8L;

C=0.730mol

1.8 L=0.406 M

C= 1.8L

0.730mol =0.406M

The student made a mistake because he did not convert a unit of volume from milliliters to liters. After all, molarity is defined as the number of moles of solute per liter of solution.

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3 years ago
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