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Nadusha1986 [10]
3 years ago
11

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

er temperature than a solution containing 5.00 g of compound B in 100. g of water. Which compound has the greater molar mass? Explain how you arrived at your answer.
Chemistry
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

Compound B.

Explanation:

The freezing point depression is a colligative property. It depends on the number of particles (moles) present in the solution.

\Delta T_{\text{f}} = K_{\text{f}} b

where b is the molal concentration

b = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}\\\\n = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{m}{M}

If m is constant (5 g), then

n \propto \dfrac{1}{M}

The compound with the greater molar mass has fewer moles and therefore fewer particles to depress the freezing point.

That must be Compound B, because Compound A has the lower freezing point.

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Answer:

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Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

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