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3 years ago

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A certain ionic compound X has a solubility in water of 40.3 g/L at 20. degrees C. Calculate the mass X of required to prepare 5

00. mL of a saturated solution of X in water at this temperature. Be sure your answer has the correct unit symbol and 3 significant digits.

1 answer:

tino4ka555 [31]3 years ago

7 0

Answer:

20.1 g

Explanation:

The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.

So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:

40.3 g/L = m/V

40.3 g/L = m/0.5L

m = 40.3 g/L * 0.5L

m = 20.1 g

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A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

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<u>Explanation:</u>

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The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

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Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of $[OH^-]$ be x

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at eq'm c-x x x

Expression of $K_b$ :

$K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}$

Since ,methyl-amine is a weak base,c>>x so $c-x\approx c$ .

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