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3 years ago

In a titration, 25.9 mL of 3.4 x10^-3 M Ba(OH)2 neutralized 16.6 mL of HCL solution. What is the molarity of the HCL solution?

1 answer:

4 0

First, We have to write the equation for neutralization:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
so, from the equation of neutralization, we can get the ratio between Ba(OH)2 and HCl. Ba(OH)2 : HCl = 1:2
- We have to get the no.of moles of Ba(OH)2 to do the neutralization as we have 25.9ml of 3.4 x 10^-3 M Ba(OH)2.
So no.of moles of Ba(OH)2 = (25.9ml/1000) * 3.4x10^-3 = 8.8 x 10^-5 mol
and when Ba(OH)2 : HCl = 1: 2
So the no.of moles of HCl = 2 * ( 8.8x10^-5) = 1.76 x 10^-4 mol

So when we have 1.76X10^-4 Mol in 16.6 ml (and we need to get it per liter)
∴ the molarity = no.of moles / mass weight
= (1.76 x 10^-4 / 16.6ml)* (1000ml/L) = 0.0106 M Hcl

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3 years ago

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From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

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Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

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Mole of C12H22O11 =?

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Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

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Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

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3 years ago

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Answer:

<u>0.150 grams</u>

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The mole ratio is:

$\dfrac{1molSnO_2}{2molH_2O}$

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