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shusha [124]
3 years ago
6

How would a solute affect the boiling point of water?

Chemistry
2 answers:
Alex787 [66]3 years ago
8 0
The answer is C. The water will boil at a higher temperature. This is because the solute is an impurity that raises the boiling point and lowers the melting point of water.
ivann1987 [24]3 years ago
6 0

Answer: c.)

Explanation: A solute raises the boiling point of water. A solution with any solvent and non-volatile solute has a higher boiling point and a lower freezing point than the pure solvent. The amount by which the boiling point increases depends on the concentration of particles but not on the identity of the solute.

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If the density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cub
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Answer:

1.35 × 10⁴ kg/m³ at 22 °C; 1.34 × 10⁴ kg/m³ at 100 °C

Explanation:

The cubic expansivity (γ) of a liquid is the fractional change in volume per unit change in temperature.

\gamma = (\frac{\Delta V }{ V_0} )(\frac{1 }{ \Delta T} )   Multiply by V₀ΔT and transpose

ΔV = γV₀ΔT  

and

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===============

<em>At 0 °C </em>

Assume you have 1 m³ of Hg

ρ = m/V     Multiply by V and transpose

m = ρV

ρ = 1.36 × 10⁴ kg/m³

m = 1.36 × 10⁴ × 1 = 1.36 × 10⁴ kg

===============

<em>At 22 °C </em>

Assume that you have 1 m³ of Hg

γ = 180 × 10⁻⁶ K⁻¹

ΔT = 22 °C – 0 °C = 22 °C

ΔV = 180 × 10⁻⁶ × 22

ΔV = 3.96 × 10⁻³ m³      Calculate volume

V = 1 + 0.00396

V = 1.00396 m³             Calculate density

ρ = 1.36 × 10⁴/1.00396

ρ = 1.35 × 10⁴ kg/m³

===============

<em>At 100 °C </em>

ΔT = 100 °C – 0 °C = 100 °C

ΔV = 180 × 10⁻⁶ × 100

ΔV = 0.0180 m³      Calculate volume

V = 1 + 0.0180

V = 1.0180 m³          Calculate density

ρ = 1.36 × 10⁴/1.0180

ρ = 1.34 × 10⁴ kg/m³

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