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Leokris [45]
3 years ago
13

A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1.

A solution is prepared by dissolving some urea ((NH2)2CO) in 550. g of Χ. This solution freezes at-0.9 °C. Calculate the mass of urea that was dissolved. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Harrizon [31]3 years ago
7 0

Answer:

27.60 g urea

Explanation:

The <em>freezing-point depression</em> is expressed by the formula:

  • ΔT= Kf * m

In this case,

  • ΔT = 5.6 - (-0.9) = 6.5 °C
  • Kf = 7.78 °C kg·mol⁻¹

m is the molality of the urea solution in X (mol urea/kg of X)

First we<u> calculate the molality</u>:

  • 6.5 °C = 7.78 °C kg·mol⁻¹ * m
  • m = 0.84 m

Now we<u> calculate the moles of ure</u>a that were dissolved:

550 g X ⇒ 550 / 1000 = 0.550 kg X

  • 0.84 m = mol Urea / 0.550 kg X
  • mol Urea = 0.46 mol

Finally we <u>calculate the mass of urea</u>, using its molecular weight:

  • 0.46 mol * 60.06 g/mol = 27.60 g urea

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Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})  

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

[CH_3COONa]=?mol  

[CH_3COOH]=0.200mol

pH = 5.00

Putting values in above equation, we get:

5=4.74+\log(\frac{[CH_3COONa]}{0.200})

[CH_3COONa]=0.364mol

To calculate the mass of sodium acetate for given number of moles, we use the equation:

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

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g

Hence, the mass of sodium acetate that must be added is 30.23 grams

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Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

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571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

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Answer:

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Explanation:

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