List three properties of water that are related to hydrogen bonding

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

7 0

3 years ago

Calculate the freezing point of the solution containing 0.105 m k2s. calculate the boiling point of the solution above. -g

Licemer1 [7]

Quantity of K2S m = 0.105 m
Number of ions i = 2(K) + 1(S) = 3
Freezing point depression constant of water Kf = 1.86
delta T = i x m x Kf = 3 x 0.105 x 1.86 = 0.586
Freezing point = 0 - 0.586 = 0.586 C
Boiling point constant of water Kb = 0.512
delta T = i x m x Kb = 3 x 0.105 x 0.512 = 0.161
Boiling point = 100 + 0.161 = 100.161 C

5 0

3 years ago

Read 2 more answers

Write a balanced chemical equation for the standard formation reaction of liquid acetic acid hch3co2.

Afina-wow [57]

The balanced chemical equation for the standard formation reaction of liquid acetic acid is given as ,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The reaction that form the products from their elements in their standard state is called formation of reaction .The acetic acid consist C , H , and O , So, determine their standard state . Carbon is graphite at 25°C and 1 atm , whereas hydrogen and oxygen are diatomic gases . Hence , we start with unbalanced reaction.

$C(gr) +H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The balanced chemical equation for the standard formation reaction of liquid acetic acid as,

$2C(gr) +2H_{2} (g) +O_{2} (g)$ → $CH_{3} COOH(l)$

The combustion of liquid acetic acid is given as,

$CH_{3} COOH(l) + 2O(g)$ → $2CO_{2}((g) +2H_{2} O(l)$ ΔH =-873

learn more about balancing chemical equation

brainly.com/question/15052184

#SPJ4

5 0

2 years ago