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mars1129 [50]
3 years ago
9

The student diluted 50.00 mL of commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-mL aliquot of t

he diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na2S2O3 solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g/mL
Determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
3 0
To get the volume of the original solution in the dilution, we consider the dilutions that were made:

from 50 mL to 250 mL:

<span>1/5 of original was present</span>

<span>from 20 mL of 250 mL:</span>

<span>1/12.5 of the diluted mixture was presetn</span>

<span>So, V=50 mL*(1/5)*(1/12.5)= 0.8 mL of original bleaching solution was present in dilution</span>

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Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

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pressure of SF_4 = 792126.36

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Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

amount of SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol

mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

mole of fractionSF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669

PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

mole of SF_4=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

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