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galben [10]
3 years ago
5

A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammo

nia. What is the concentration of the original ammonia solution?
Chemistry
1 answer:
Kryger [21]3 years ago
8 0

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

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The law states that "equal volumes of all gases at the same temperature and pressure contains equal number of molecules or moles".

The law describes the behavior of gases when involve in chemical reactions. It enables one to change over at will in any statement about gases from volumes to molecules and vice versa.

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Heat of reaction (i.e enthalpy of reaction) is the quantity of heat that is required to be added or removed when a chemical reaction is taken place in order to maintain all of the compounds present at the same temperature.

The formula used to calculate the heat of the reaction can be expressed as follows:

Q = mcΔT

where:

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From the information given:

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∴

The change in temperature i.e. ΔT = T₂ - T₁

ΔT = 91.5° C - 25° C

ΔT = 66.5° C

The number of moles of CuSO₄ = 1.00 mol/dm³ × 50.0 cm³

\mathbf{= (1 \times \dfrac{50}{1000})\ moles}

= 0.05 moles

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Then;

Using the relation:

\mathbf{number \ of \ moles = \dfrac{mass}{molar \ mass}}}

By crossing multiplying;

mass of CuSO₄ = number of moles of CuSO₄ ×  molar mass of CuSO₄

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∴

Using the formula from above:

Q = mcΔT

Q = 7.9805 g × 4.18 kJ/g °C × 66.5° C

Q = 2218.34 kJ

Therefore, we can conclude that the heat of the reaction is 2218.34 kJ

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