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lesya692 [45]
2 years ago
13

A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo

iling point of a solution made of 11. g of ammonium chloride dissolved in 400 g of X.
Chemistry
1 answer:
Goshia [24]2 years ago
8 0

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

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If a volume of gas 177.6mL was collected at a temperature of 25.8C and the pressure is 799.7 torr, what is the original concentr
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Answer:

The original concentration of the acid was 0.605 M

Explanation:

Step 1: Data given

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Volume of acid needed to react = 12.6 mL = 0.0126 L

Step 2: Calculate moles

p*V = n*R*T

n = (p*V)/(R*T)

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⇒with V = the volume of the gas = 177.6 mL = 0.1776 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

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n = 0.007618 moles

Step 3: Calculate original concentration

We need 0.007618 moles of acid to react with the same amount of moles gas

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