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BARSIC [14]
3 years ago
11

A student prepares 150. mL of a 2.00 M Ba(OH)2 aqueous solution in lab. What mass of Ba(OH)2 is fully dissolved in the solution?

Chemistry
1 answer:
meriva3 years ago
3 0

Answer: 51.4g

Explanation:

Given that:

Amount of moles of Ba(OH)2 (n) = ?

Volume of Ba(OH)2 solution (v) = 150.0mL

[Convert 150.0mL to liters

If 1000 mL = 1L

150.0mL = 150.0/1000 = 0.150L]

Concentration of Ba(OH)2 solution (c) = 2.00M

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 2.00M x 0.150L

n = 0.3 mole

Now given that,

Amount of moles of Ba(OH)2 (n) = 0.3

Mass of Ba(OH)2 in grams (m) = ?

For molar mass of Ba(OH)2, use the molar masses:

Barium, Ba = 137.3g;

Oxygen, O = 16g;

Hydrogen, H = 1g

Ba(OH)2 = 137.3g + [(16g + 1g) x 2]

= 137.3g + [17g x 2]

= 137.3g + 34g

= 171.3 g/mol

Since, amount of moles = mass in grams / molar mass

0.3 mole= m / 171.3g/mol

m = 0.3 mole x 171.3g/mol

m = 51.39g

[Round the 51.39g to the nearest tenth, so it becomes 51.4g]

Thus, the mass of Ba(OH)2 fully dissolved in the solution is 51.4 grams

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2 years ago
Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.
dexar [7]

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Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

 HClO +H2O ⇒ ClO- + H3O+

I  0.265                0        0

C  -x                    +x     +x

E  0.265-x          +x      +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

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8 0
3 years ago
How many grams of NaCl (molecular weight = 58.4 g mole-1) would you dissolve in water to make a total volume of 500 mL of soluti
kolezko [41]

Given :

Volume , V = 500 mL .

Molarity , M = 0.5 M .

Molecular mass of NaCl is 58.4\ g/mole .

To Find :

How many grams of NaCl is required .

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Let , NaCl required is x gram .

Molarity is given by :

M=\dfrac{\text{Number of moles}}{\text{Volume (in liters) }}\\\\M=\dfrac{m}{M\times V}\\\\0.5=\dfrac{x}{58.4\times 0.5}\\\\x=0.5^2\times 58.4\ g\\\\x=14.6\ g

Hence , this is the required solution.

6 0
3 years ago
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