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harkovskaia [24]
3 years ago
12

A student conducts an experiment to determine the effect of adding salt on the boiling temperature of water. the results are sho

wn below.
Chemistry
1 answer:
Amanda [17]3 years ago
5 0

A student conducts an experiment to determine the effect of adding salt on the boiling temperature of water. The results are shown below.

S #ofsalt BoilTemp

S1 0g 100.0°C

S2 2g 102.3°C

S3 5g 104.8°C

S4 10g 107.5°C

Which is a value that should remain constant for all samples in this experiment?

a. Amount of salt added

b. Volume of water

c. Boiling Temperature

d. Time required for water to boil


On adding solute to specific volume of solvent, the vapor pressure of the solvent decrease as the solute added is salt which is non volatile solute. Thus, it increases the boiling point furthur on adding the solute.

The amount of salt added are 51 g, 52 g, 53 g, 54 g so the concentration of solute is changing.

Temperature of boiling of solution changes on addition of solute to solution.

The time required for the solution to boil varies with amoutn of solute added, as the vapor pressure of solution decreases.

Thus, the voume of water will remain contant as only amount of salt added varies.


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Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
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Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

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Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

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Number of moles of H+=Molarity multiply by volume of given acid

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Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

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