What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?
1 answer:
Complete Question
Magnesium sulfate forms a hydrate with the formula . What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of
is 120.4 g/mol; H20 = 18 g/mol.
Answer:
The value is of
Explanation:
From the question we are told that
The volume of toluene is
The mass of anhydrous magnesium sulfate is
The formula of the hydrate is
The molar mass of is
From the formula given we see that
1 mole of wil remove 7 moles of
to for the given formula
Hence
120.4 g (1 mole) will remove 7 moles (7 * 18 g = 126 g ) of to for the given formula
Therefore 1 g of x g of
So
=>
From our calculation we obtained that
1 g of will remove
of
Then
of
will remove z g of
of
So
=>
=> of
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Answer:
The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml
Explanation:
Here we have the reaction of AgNO₃ and NaCl as follows;
AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)
Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,
Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains
0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃
Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;
0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL
Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.