1-What is volume in Liter of 10.3 moles of o2 gas at 25 ºC and 2 atm of pressure

Question

1-What is volume in Liter of 10.3 moles of o2 gas at 25 ºC and 2 atm of pressure ?

2- What is the molarity of an 85.0 mL ethanol (C2H5OH) soulation contaning 1.77g of ethanol?

3- How many grams of potassium dichromate ( K2Cr2O7) are required to prepare a 250 mL of a 2.16 M soulation ?

4- A concentraed soulation of NaOH is 50% NaOH by mass ( meaing grams of solute to grams of soulation ) and has a density of 1.54 g/mL. if 25.0 mL of this soulation is diluated to final volume of 725. mL what is the molarity of the resulting soulation ?

5- A soulation prepared by dissolving 0.300 g of an unknow nonvolatile solute in 30.0 g of carbon tertachlortride has a boiling point that is 0.392 ºC higher than that pure CCL4. the boliling point elevation constant for CCL4 is +5.02º/m. what is the moleculat weight of this solute?

(SHOW WORKS PLEASE )

Explanation / Answer

1- Using,

V = nRT/P

with,

n = 10.3 mol

R = gas constant

T = 25 + 273 = 298 K

P = 2 atm

We get,

Volume of O2 gas (V) = 10.3 x 0.08205 x 298/2 = 126 L

2- Molarity = moles/L of solution

So, molarity of ethanol solution = 1.77 g/46.07 g/mol x 0.085 L = 0.452 M

3- Molarity = grams of solute/molar mass x L of solution

So, grams of K2Cr2O7 required = 2.16 M x 0.250 L x 294.185 g/mol = 158.86 g

4- 50% NaOH (w/w) has = 50 g NaOH in 50 g of solvent.

moles of NaOH = 50 g/40 g/mol = 1.25 mols

Volume of solution = 100 g/1.54 g/ml = 64.93 ml

Molarity of original solution = 1.25 mols/0.06493 L = 19.25 M

Molarity of diluted solution = 19.25 M x 25 ml/725 ml = 0.664 M

5- Elevation in boiling point

dTb = i.Kb.m

with,

dTb = 0.392 oC

i = 1

Kb = 5.03 oC/m

m = 0.3 g/molar mass x 0.03 kg

we get,

molar mass of unknown solute = 1 x 5.03 x 0.3/0.03 x 0.392 = 128.316 g/mol

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