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Serga [27]
3 years ago
5

Determine the molality of an aqueous solution that is 15.5 percent urea by mass.

Chemistry
1 answer:
DaniilM [7]3 years ago
3 0

15.5% by mass is equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. <span>

<span>we know that molality = moles solute / kg solvent

<span>moles solute = 155 g / 60 g/mol = 2.58 moles urea

</span></span></span>

Since there are 155 g urea in 1000g solution, hence the solvent is 845 g or 0.845 kg

So:<span>
<span>molality = 2.58 / 0.845 = 3.06 m</span></span>

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A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equa
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<u>Answer:</u> The identity of the unknown acid is butanoic acid or ascorbic acid.

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0.570M=\frac{\text{Moles of NaOH}\times 1000}{39.55}\\\\\text{Moles of NaOH}=\frac{0.570\times 39.55}{1000}=0.0225mol

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By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HX

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2NaOH+H_2X\rightarrow 2NaX+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of diprotic acid

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To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

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Molar mass of butanoic acid = 88 g/mol

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\text{Moles of butanoic acid}=\frac{2.002g}{88g/mol}=0.02275mol

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Molar mass of L-tartaric acid = 150 g/mol

Putting values in above equation, we get:

\text{Moles of L-tartaric acid}=\frac{2.002g}{150g/mol}=0.0133mol

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\text{Moles of ascorbic acid}=\frac{2.002g}{176g/mol}=0.01137mol

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Hence, the identity of the unknown acid is butanoic acid or ascorbic acid.

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