Chemistry

5).
Inthe determination of the Molar Volume of a Gas experiment a volume of 35.0 mLof gas is collected at 24.0°C and a pressure of 762 mm Hg . Vapor pressure ofwater at 24.0°Cis 19.83 mm Hg. Show calculations for a and b
a)What is the pressure of dry gas?
b)How many moles of dry gas are present?

a)
Pressure of dry gas = 762 - 19.83 = 742.17 mm Hg

b)
PV = nRT
(742.17/760) x (35/1000) = n x 0.082 x (273 + 24)
No. of moles of dry gas, n = 0.00140 mol


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6)A 1.00 mL sample of NaOH was standardized using 1.50 mL of 0.0970 M potassiumhydrogen phthalate (KHP). The standardized solution was used to titrate 0.250mL of the vinegar. The volume of NaOH needed was 4.510 mL. Density of thevinegar = 1.005g/mL. Molar mass of CH3COOH= 60.05g.
Find the concentration of acetic acid in the vinegar in:
a)moles per liter. (Show calculations)
b)grams per liter(Show calculations)
c)% by weight (Show calculations)

a)
NaOH + KHP →NaKP + H2O
No. of moles of KHP = 0.0970 x (1.50/1000) = 0.0001455 mol
No. of moles of NaOH = 0.0001455 mol
Molarity of NaOH = 0.0001455 / (1.00/1000) = 0.1455 mol/L

NaOH + CH3COOH → CH3COONa + H2O
No. of moles of NaOH = 0.1455 x (4.510/1000) = 0.0006562 mol
No. of moles of CH3COOH = 0.0006562 mol
Concentration of CH3COOH = 0.0006562 / (0.25/1000) = 2.625mol/L

b)
Concentration of CH3COOH = 2.625 * 60.05 = 157.6 g/L

c)
Consider 1 L (1000 mL) of the vinegar.
Total mass of the vinegar = 1000 x 1.005 = 1005 g
Mass of CH3COOH = 157.6 g
% by weight of CH3COOH = (157.6/1005) x 100% = 15.68%

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2)
100 mL of 0.750 M NaOH solution is neutralized with1 M phosphoric acid H3PO4.
Write balanced chemical equation. How many mL ofphosphoric acid will be needed?
Show calculations

H3PO4 + 3NaOH → Na3PO4 + 3H2O
No. of moles of NaOH = 0.75 x (100/1000) = 0.075 mol
No. of moles of H3PO4 = 0.075 x (1/3) = 0.025 mol
Volume of H3PO4 = 0.025/1 = 0.025 L = 25 mL