Chemistry, 29.06.2019 23:30 quan91

The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van't hoff factor for this solute if the kb for water is 0.512°c/m. 2.7 1.8 2.3 3.0 3.6

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Answer from: ameera1973

The value of the van't Hoff factor for the given solute is 2.7

Option (A) 2.7 is correct.

What is van't Hoff factor?

The Van't Hoff factor is a measure of a solute's effect on colligative qualities such boiling point elevation, osmotic pressure, and relative vapor pressure reduction.

Given,

Molar mass is 132.15 g/mol

Boiling point elevation is 102.5 °C

kb for water is 0.512°c/m

Molarity of solution is 1.83 m

The value of van't Hoff factor is: 2.66 or 2.7 (approx)

Now, from the solution of colligative properties to calculate elevation in boiling point.

$\bold{\Delta Tb = ikbm}$

where Δ = elevation of boiling point (102.5)

$\bold{102.5\times i\times 0.512 \times 1.83 }$

$\bold{i = 2.66 = 2.7 }$

Thus, The value is 2.7 option (A) is correct.

Learn more about van't Hoff factor, here:

link

Answer from: desi3187

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: Δ $T_{b}$ =i $K_{b}$ .m, where Δ $T_{b}$ = elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

$K_{b}$ = Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i= $\frac{2.5}{0.512 X 1.83}$

i=2.66= 2.7 (approx.)

Answer from: Quest

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explanation:

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Answer from: Quest

answer: e. sour taste

explanation: acids are those substances which dissociates in water to give $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

$hcl\rightarrow h^++cl^-$

bases are those substances which dissociates in water to give $oh^-$ ions or accepts $h^+$ ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

$naoh\rightarrow na^++oh^-$

$nh_3+h^+\rightarrow nh_4^+$

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The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. de...

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