Name Chem 1120 Homework3 1. Rubbing alcohol contains 585 g isopropanol (CH-OH) p

Question

Name Chem 1120 Homework3 1. Rubbing alcohol contains 585 g isopropanol (CH-OH) per liter (aqueous). Calculate the molarity. The density of water is 1.00 g/mL 2. What volume of 0.25 M HCI solution must be diluted to prepare 1.00 L of 0.040 M HCI? 3. Calculate the sodium ion concentration when 70.0 mL of 3.0 M sodium carbonate is added to 30.0 ml of 1.0 M sodium bicarbonate. 4. Write equations showing the ions present after the following strong electrolytes are dissolved in water a. HNO b. Na,SO C. Al(NOs d. SrBr e. KCIO f. NH.Br

Explanation / Answer

1)

Mass of Isopropanol = 585 g

molar mass of isopropanol(H3CCHOHCH3) = 60.1g/mol

Molaity = mass/molar mass*1000/Volume of solution

Molarity = 585/60.1*1000/1000

Molarity = 9.734 mol/L

2)

we Have the formula M1V1 = M2V2

M1 = initial concentration = 0.25 M

V1 = initial volume = ?

M2 = final concentration = 0.04 M

V2 = final volume = 1L

so, 0.25 * V1 = 0.04 * 1

V1 = 0.16 L

So the volume of water required to dilute the solution = 1 - 0.16 = 0.84 L

3)

Number of moles of Na2CO3 = 70 * 3 = 210 mmole

Number of moles of NaHCO3 = 30 * 1 = 30 mmole

Na2CO3 ------> 2Na+ + Co32-

1 mole of Na2CO3 produces 2 moles of Na+ ion

210 mmoles of Na2CO3 produces 2 * 210 mmoles of Na+ ion

                             = 420 mmoles

NaHCO3 -----> Na+ + HCO3-

1 mole of NaHCO3 produces 1 mole of Na+ ions

30 mmol of NaHCO3 produces 30 mmol of Na+ ion

Total number of moles of Na+ ion = 420 + 30 = 450 mmol

Concentration of Na+ ion = 450/(70+30) = 4.5 mol/L

4)

HNO3 ------> H+ + NO3-

Na2SO4 -----> 2 Na+ + SO4 2-

Al(NO3)3 -----> Al+3 + 3 NO3 -

SrBr2 ------> Sr+2 + 2 Br-

KClO4 ------> K+ + ClO4-

NH4Br ------> NH4+ + Br-

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