Why are concentration units of mole fraction used in some colligative properties calculations?

1 answer:

4 0

The answer is d. mole fraction is not affected by changes in temperature. The colligative properties is only depend on the number of the solute in the solvent.

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calculate the number of grams of CH3COONa * 3H2O (sodium acetate tri-hydrate) needed to make 250.0 mL of a CH3COOH (acetic acid)

anastassius [24]

Answer:

10.88 g

Explanation:

We have:

[CH₃COOH] = 0.10 M

pH = 5.25

Ka = 1.80x10⁻⁵

V = 250.0 mL = 0.250 L

$M_{CH_{3}COONa*3H_{2}O} = molar \thinspace mass = 136 g/mol$

The pH of the buffer solution is:

$pH = pKa + log(\frac{[CH_{3}COONa*3H_{2}O]}{[CH_{3}COOH]})$ (1)

By solving equation (1) for [CH₃COONa*3H₂O] we have:

$log [CH_{3}COONa*3H_{2}O] = pH - pKa + log [CH_{3}COOH]$

$log [CH_{3}COONa*3H_{2}O] = 5.25 - (-log(1.80 \cdot 10^{-5})) + log (0.10) = -0.495$

$[CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M$

Hence, the mass of the sodium acetate tri-hydrate is:

$m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g$

Therefore, the number of grams of CH₃COONa*3H₂O needed to make an acetic acid/sodium acetate tri-hydrate buffer solution is 10.88 g.

I hope it helps you!

4 0

3 years ago

PLEASE HELP!! <br> this is on USAtestprep <br> a)<br> b)<br> c)<br> d)

kaheart [24]

The answer is D!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3 0

3 years ago

CaC2 + 2H2O → C2H2 + Ca(OH)2 If 2.8 moles of CaC2 are consumed in this reaction, how many grams of H2O are needed?

Strike441 [17]

Answer:

Cac2 is a answer please mark me brainliest

8 0

3 years ago

A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1,150 g of water at 25.0°C. If the final temperature of the system

AleksAgata [21]

To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:

$-m\cdot c_e\cdot \Delta T = m_w\cdot c_e_w\cdot \Delta T_w$

Clearing the specific heat of the unknown substance:

$c_e = \frac{m_w\cdot c_e_w\cdot \Delta T_w}{m\cdot \Delta T} = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot (37.1 - 25.0)^\circ C}{75\ g\cdot (37.1 - 96.5)^\circ C}$

$c_e = -\frac{1\ 150\ g\cdot 4.184\frac{J}{g\cdot ^\circ C}\cdot 12.1^\circ C}{75\ g\cdot (-59.4)^\circ C} = \bf 13.07\frac{J}{g\cdot ^\circ C}$