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LenKa [72]
3 years ago
8

A student used a DCP solution (standardized to 9.98x10-4M) to analyze a sample of extract from 2.50g of solid food. The titratio

n required 22.49mL of DCP. 87.00% of all the ascorbic acid in the food was collected in the extract. (MM Ascorbic Acid = 176.124 g/mol)
What mass of food (in grams) would be required to attain the RDA of ascorbic acid (60 mg)
Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

Given that the molarity of DCP equals 9.98x10-4M. And it used to analyze a sample of extract from 2.50g of solid food. The titration required 22.49mL of DCP. 87.00% of all the ascorbic acid in the food was collected in the extract.

 Amount of solid = 60 mg AA (100mg/87mg)(1g/1000mg)(1 mol AA/176.124 gAA)(1mol DPC/1mol AA)( 1L DCP/9.98x10-4mol DPC)(2.50g solid/0.02249 L DPC) = 43.6 g solid.

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Water has a vapor pressure of 23.8 mm Hg at 25°C and a heat of vaporization of 40.657 kJ/mol. Using the Clausius-Clapeyron equat
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Answer:

P = 559.553 mmHg

Explanation:

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∴ P1 = 23.8 mmHg = 3.173 KPa

∴ T1 = 25°C ≅ 298 K

∴ ΔHv = 40.657 KJ/mol

∴ R = 8.314 E-3 KJ/K.mol

∴ T2 = 96°C ≅ 369 K

⇒ Ln P2/P1 = - (40.657 KJ/mol/8.314 E-3 KJ/K,mol) [(1/369 K) - (1/298 K) ]

⇒ Ln P2/P1 = - (4890.185 K) [ - 6.457 E-4 K-1 ]

⇒ Ln P2/P1 = 3.1575

⇒ P2/P1 = 23.511

⇒ P2 = (23.511)(3.173 KPa)

⇒ P2 = 74.601 KPa = 559.553 mmHg

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A particular flask has a mass of 17.4916 g when empty. When filled with ordinary water at 20.0°c (density = 0.9982 g/ml), the ma
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The mass of the empty flask is 17.4916 g. Now after feeling the ordinary water the mass of the flask is 43.9616 g. Thus the change of weight due to addition of ordinary water is (43.9616 - 17.4916) = 26.47 g.

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