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2 years ago

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An aqueous solution of calcium hydroxide is standardized by titration with a 0.120 M solution of hydrobromic acid. If 16.5 mL of

base are required to neutralize 27.5 mL of the acid, what is the molarity of the calcium hydroxide solution?

1 answer:

Artist 52 [7]2 years ago

7 0

<u>Answer:</u> The molarity of calcium hydroxide in the solution is 0.1 M

<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HBr$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL$

Putting values in above equation, we get:

$1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M$

Hence, the molarity of $Ca(OH)_2$ in the solution is 0.1 M.

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The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

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2 years ago

If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?

bija089 [108]

<h3>Molar mass:-</h3>

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3 years ago

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Answer:

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<span>6.38x10^-2 moles
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