Chemistry, 22.01.2022 06:00, cameronbeaugh

Lodine dissolved in water, but is more soluble in a solvent such as carbon tetrachloride. This reaction can be summarized by:
I2 (aq) = I2 (CC14)
K=85.0 at 25.0°C
If you dissolve 0.0340 g 12 in 100.0 mL of water and then add it to a vessel containing
25.0 mL CCl4 and shake it (for long enough that you can
assume equilibrium has been reached), how much I (what mass) has moved into the
CCl4?

Answers: 3

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answered: kayranicole1

Carbon tetrachloride is nonpolar.

answered: Guest

The answer to your question really depends on three things: how much water you have, how cold it is to start out, and how cold the things around it are. water actually freezes when it gets to 32 degrees fahrenheit (0 degrees celsius), but the time it takes to get there may be different. let's start with the first. if you take two glasses, and fill one with a tiny bit of water, and the other about halfway, then put them both in the freezer, the one with less water will freeze first (you can try this at home, but i recommend using plastic cups and not glass ones). now let's move on to the second part. let's say you have two glasses, and you fill one with really cold water that has been in the refrigerator, and the other with really hot water from the sink. if you put both of them in the freezer, the one that started out colder will freeze first. for the third part, let's imagine that you have two glasses with the same amount of water in them, and the water is at the same temperature. imagine putting one outside on a really really cold day in georgia, and having a friend in alaska put one outside on the same day. since it would be so much colder in alaska, the glass of water there would freeze before yours. so, if you took a tiny bit of really cold water in a glass, and put it outside on a cold day in alaska, it would freeze a lot faster than a big glass of hot water outside on a cold day in georgia.

answered: Guest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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