All categories

3 years ago

Which will have the lowest boiling point?

Chemistry

2 answers:

In-s [12.5K]3 years ago

8 0

Some of the solutions exhibit colligative properties. These properties depend on the amount of

solute dissolved in a solvent. These properties include freezing point depression, boiling

point elevation, osmotic pressure and vapor pressure lowering. From the given choices, the correct answer is the first option. Pure water will have the lowest boiling point because there are no solute particles to change the boiling point of water while other options will have an elevated boiling point because of the solute.

Read more on Brainly.com - brainly.com/question/1437052#readmore

Zina [86]3 years ago

4 0

Answer: A. pure water

Explanation:

$\Delta T_b=i\times k_f\times m$

$\Delta T_b$ = change in boiling point

i = Van'T Hoff factor

$k_f$ = freezing point constant

m = molality

1. For pure water

, i= 1 as it is a non electrolyte and does not dissociate.

2. For $0.5M NaCl$

$NaCl\rightarrow Na^+Cl^-$

, i= 2 , as it is a electrolyte and dissociate to give 2 ions and concentration will be (0.5+0.5)=1M[/tex]

3. For $1M NaCl$

$NaCl\rightarrow Na^+Cl^-$

, i= 2 , as it is a electrolyte and dissociate to give 2 ions and concentration will be (1+1)=2M[/tex]

4. For $2M NaCl$

$NaCl\rightarrow Na^+Cl^-$

, i= 2 , as it is a electrolyte and dissociate to give 2 ions and concentration will be (2+2)=4M[/tex]

Thus as vant hoff factor is lowest for pure water and thus the boiling point will be lowest.

You might be interested in

Select the person responsible for the following: printing press

Alinara [238K]

That would be Johannes Gutenberg a German inventor from the 1400's

4 0

3 years ago

Argon atom can exist freely in nature, why

lana [24]

Answer:

Argon can exist freely in nature because it has a full octet of electrons the way its found in the nature is the same way its found in periodic table of element in vast amouts of stabilization.

7 0

3 years ago

what is the chemical reaction in which the exchange of electropositive and electronegative ions occurs in aqueous solution

defon

Answer:

chemical reaction are the processes in which new substances with new properties of from are called as chemical reaction ex double displacement reaction

Explanation:

7 0

3 years ago

Read 2 more answers

Which of the following compounds are polar: CH2Cl2, HBr?

Olegator [25]

Answer : HBr polar

please add me to brainalist

8 0

3 years ago

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

7 0

3 years ago