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shtirl [24]
3 years ago
5

Your lab group forgot to record the initial volume measurement on the buret and assumed later that it started at 0.00 mL when th

e actual starting buret reading was 2.00 mL. Would your final calculated molarity of the unknown acid be higher, lower or equal to the actual concentration
Chemistry
1 answer:
kherson [118]3 years ago
4 0

Answer:

Calculated molarity of acid is higher than the actual concentration.

Explanation:

For an acid-base titration, the following equation is obeyed for calculation-

                                   C_{acid}.V_{acid}=C_{base}.V_{base}

where, C represents concentration and V represents volume.

So, C_{acid}=\frac{C_{base}.V_{base}}{V_{acid}}

Here, C_{base} and V_{acid} are constants.

Actual volume of base added from burette is lesser than the volume used (based on assumption) for calculation of concentration of acid.

So, actual value of V_{base} is lesser. Hence calculated concentration or molarity of acid is higher than the actual concentration.

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What is the molarity of solution obtained when 5.71 g of sodium carbonate-10-water is dissolved in water and made up to 250.0 cm
disa [49]

We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.

The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O)  is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³

The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.

Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.

5.71 grams of Na₂CO₃.10 H₂O is equal to \frac{5.71}{286}= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.

Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is \frac{0.0199 X 1000}{250} = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³


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