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3 years ago

A student is instructed to make 1 L of a 2.0 M solution of CaCl2 using dry salt. How should he do this?

1 answer:

4 0

The student should follow following steps to make 1 L of 2.0 M CaCl₂.

1. First he should calculate the number of moles of 2.0 M CaCl₂ in 1 L solution.

Molarity of the solution = 2.0 M
Volume of solution which should be prepared = 1 L

Molarity = number of moles / volume of the solution

Hence, number of moles in 1 L = 2 mol

2. Find out the mass of dry CaCl₂ in 2 moles.

moles = mass / molar mass

Moles of CaCl₂ = 2 mol
Molar mass of CaCl₂ = 110.98 g/mol

Hence, mass of CaCl₂ = 2 mol x 110.98 g/mol
= 221.96 g

3. Weigh the mass accurately

4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and finally wash the funnel and watch glass with de-ionized water. That water also should be added into the volumetric flask.

5. Then add some de-ionized water into the volumetric flask and swirl well until all salt are dissolved.

6. Then top up to mark of the volumetric flask carefully.

7. As the final step prepared solution should be labelled.

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After a recrystallization, a pure substance will ideally appear as a network of __________. If this is not the case, it may be w

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Answer:

Large crystals

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Explanation:

Recrystallization is a procedure used to purify an impure compound in a solvent. The concept used here is that the solubility of most of the solids increases with increase in temperature.

Recrystallization is also called as method of fractional crystallization.

After a recrystallization, a pure substance will ideally appear as a network of large crystals. If this is not the case, it may be worthwhile to reheat the flask and allow the contents to cool more slowly.

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3 years ago

Assume that the batteries in a flashlight are rechargeable. What energy conversion would have to take place in order to recharge

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Explanation:

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4 years ago

Assuming the boiling point increased by 2 Â°C, what is the approximate molality of NaCl when the video ends? Kbp(water) = 0.512

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Answer : The molality of NaCl is, 1.95 mol/kg

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = change in boiling point = $2^oC$

$k_b$ = boiling point constant for water = $0.512^oC/m$

m = molality

i = Van't Hoff factor = 2 (for electrolyte)

The dissociation $NaCl$ will be,

$NaCl\rightarrow Na^++Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Na^++Cl^{-}$ = 1 + 1 = 2

Now put all the given values in the above formula, we get:

$2^oC=2\times (0.512^oC/m)\times m$

$m=1.95mol/kg$

Therefore, the molality of NaCl is, 1.95 mol/kg

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