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sashaice [31]
2 years ago
11

How do you figure out molality

Chemistry
1 answer:
hodyreva [135]2 years ago
3 0
*The molality of a solution is calculated by taking the moles of solute and dividing by
the kilograms of solvent* Basically if we had 1.00 mole of sucrose (it's about 342 3 grams) and
proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar
water. We keep adding water, dissolving and stirring until all the solid was gone. We
then made sure everything was well-mixed.
What would be the molality of this solution? Notice that my one liter of water weighs
1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter).
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A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo
Goshia [24]

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

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8 0
3 years ago
Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol
Sonbull [250]

<u>Answer:</u> The value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

                N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>            1.30       1.65

<u>At eqllm:</u>       1.30-x    1.65-3x             2x

Evaluating the value of 'x'

\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol

The expression of K_c for above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

Equilibrium moles of nitrogen gas = (1.30-x)=(1.30-0.05)=1.25mol

Equilibrium moles of hydrogen gas = (1.65-x)=(1.65-0.05)=1.60mol

Putting values in above expression, we get:

K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}

Calculating the K_c' for the given chemical equation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2

Hence, the value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

8 0
3 years ago
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