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Neko [114]
3 years ago
10

Which of these solutes raises the boiling point of water the most?

Chemistry
2 answers:
Vlad [161]3 years ago
7 0
I think the answer would be Ionic sodium phosphate (Na3PO4) because it has the greatest boiling point elevation.
never [62]3 years ago
7 0

Answer: ionic sodium phosphate (Na_3PO_4)

Explanation:

The formula used:

\Delta T_b=i\times k_b\times m

\Delta T_b= change in boiling point

i = Van'T Hoff factor  

k_b = boiling point constant

m = molality

1.  For LiCl

LiCl\rightarrow Li^{+}+Cl^{-}  

i= 2 as it is ionic and dissociate to give 2 ions.

2. For NaCl

NaCl\rightarrow Na^{+}+Cl^{-}  

i= 2 as it is ionic and dissociate to give 2 ions.

3. For C_{12}H_{22}O_{11}

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

4. For Na_3PO_{4}

Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}  

, i= 4 as it is ionic and dissociate to give 4 ions.

5.  For MgBr_{2}

MgBr_2\rightarrow Mg^{2+}+2Br^-  

, i= 3 as it is ionic and dissociate to give 3 ions.

Thus as vant hoff factor is highest for Na_3PO_{4} , the boiling point will be raised most.

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5 0
3 years ago
A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0
3 years ago
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