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2 years ago

Al wonders if adding salt to water will affect how long it takes the water to

Chemistry

1 answer:

love history [14]2 years ago

6 0

The hypothesis is that salt water freezes faster than fresh water.

The dependent variable is time taken for ice to appear.

The independent variable is presence or absence of salt

The constants are the amount of water in each tray, freezing conditions and length of time of exposure to freezing condition.

The control group is the tray to which salt was not added

The experimental group is the tray to which salt was added

The presence of solutes in a solution causes the freezing point depression.

A solution is made up of a solute and a solvent. In the presence of a solute, the freezing point of a pure solvent is decreased. This is because freezing point is a colligative property.

Colligative properties depend on the amount of solute present.

Hence, the pure water freezes faster (ice begin to appear earlier) than the salt water.

The hypothesis put forward in this experiment was found to be invalid by the experiment.

For more about colligative properties, see

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Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .

Hence, $K_{sp} = x[Br^{-}]^{2}_{o}$

$4.67 \times 10^{-6} = x \times (0.10)^{2}$

x = $4.67 \times 10^{-4}$ M

Thus, we can conclude that molar solubility of $PbBr_{2}$ is $4.67 \times 10^{-4}$ M.

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