Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= 
× 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
Answer:
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Explanation:
As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.
Hence, amount of diethyl ether present will be calculated as follows.
(100ml - 1.468 ml)
= 98.532 ml
So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.
Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.
Amount of water = 
= 0.3427 ml
Now, when magnesium dissolves in water then the reaction will be as follows.
Molar mass of Mg = 24.305 g
Molar mass of 
= 18 g
Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.
Amount of Mg = 
= 0.462 g
