1. Bartender 3 0 64 Grams =
2. Bartender 3 0 64 Grams Equal

1. Pennyweights to Grams table Start Increments Increment: 1000 Increment: 100 Increment: 20 Increment: 10 Increment: 5 Increment: 2 Increment: 1 Increment: 0.1 Increment: 0.01 Increment: 0.001 Fractional: 1/64 Fractional: 1/32 Fractional: 1/16 Fractional: 1/8 Fractional: 1/4 Fractional: 1/2.
2. Density/grams per cm 3 Element Name Density/grams per cm 3; Actinium 10.07 Mendelevium Aluminum 2.702 Mercury 13.546 Americium 13.67 Molybdenum 10.2 Antimony 6.684 Moscovium Argon 0.001784 Neodymium 7.0 Arsenic 5.776 Neon 0.0009 Astatine 7 Neptunium 20.45 Barium 3.59 Nickel 8.91 Berkelium 14.79 Nihonium Beryllium 1.848 Niobium 8.57 Bismuth 9.
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Grams to Ounces table Start Increments Increment: 1000 Increment: 100 Increment: 20 Increment: 10 Increment: 5 Increment: 2 Increment: 1 Increment: 0.1 Increment: 0.01 Increment: 0.001 Fractional: 1/64 Fractional: 1/32 Fractional: 1/16 Fractional: 1/8 Fractional: 1/4 Fractional: 1/2.

Molarity Problems
#11 - 25

Problem #11: What volume (in mL) of 12.0 M HCl is needed to contain 3.00 moles of HCl?

Solution:

12.0 M = 3.00 mol / x

x = 0.250 L

This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value:

0.250 L x (1000 mL / L) = 250. mL (note use of explicit decimal point to create three sig figs)

Problem #12: How many grams of Ca(OH)₂ are needed to make 100.0 mL of 0.250 M solution?

Solution:

(0.250 mol L¯¹) (0.100 L) = x / 74.0918 g mol¯¹

x = (0.250 mol L¯¹) (0.100 L) (74.0918 g mol¯¹)

x = 1.85 g (to three sig figs)

Problem #13: What is the molarity of a solution made by dissolving 20.0 g of H₃PO₄ in 50.0 mL of solution?

Solution:

(x) (0.0500 L) = 20.0 g / 97.9937 g mol¯¹

(x) (0.0500 L) = 0.204094753 mol

x = 4.08 M

Problem #14: What weight (in grams) of KCl is there in 2.50 liters of 0.500 M KCl solution?

Solution:

(0.500 mol L¯¹) (2.50 L) = x / 74.551 g mol¯¹

x = 93.2 g

Problem #15: What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?

Solution:

(x) (0.2500 L) = 12.0 g / 39.9969 g mol¯¹<--- note use of L in set up, not mLx = 1.20 M

Problem #16: Determine the molarity of these solutions:

a) 4.67 moles of Li₂SO₃ dissolved to make 2.04 liters of solution.
b) 0.629 moles of Al₂O₃ to make 1.500 liters of solution.
c) 4.783 grams of Na₂CO₃ to make 10.00 liters of solution.
d) 0.897 grams of (NH₄)₂CO₃ to make 250 mL of solution.
e) 0.0348 grams of PbCl₂ to form 45.0 mL of solution.

Solution set-ups:

a) x = 4.67 mol / 2.04 L
b) x = 0.629 mol / 1.500 L
c) (x) (10.00 L) = 4.783 g / 106.0 g mol¯¹
d) (x) (0.250 L) = 0.897 g / 96.09 g mol¯¹
e) (x) (0.0450 L) = 0.0348 g / 278.1 g mol¯¹

Problem #17: Determine the number of moles of solute to prepare these solutions:

a) 2.35 liters of a 2.00 M Cu(NO₃)₂ solution.
b) 16.00 mL of a 0.415-molar Pb(NO₃)₂ solution.
c) 3.00 L of a 0.500 M MgCO₃ solution.
d) 6.20 L of a 3.76-molar Na₂O solution.

Solution set-ups:

a) x = (2.00 mol L¯¹) (2.35 L)
b) x = (0.415 mol L¯¹) (0.01600 L)
c) x = (0.500 mol L¯¹) (3.00 L)
d) x = (3.76 mol L¯¹) (6.20 L)

Comment: the technique used is this:

MV = moles of solute

This particular variation of the molarity equation occurs quite a bit in certain parts of the acid base unit.

Problem #18: Determine the grams of solute to prepare these solutions:

a) 0.289 liters of a 0.00300 M Cu(NO₃)₂ solution.
b) 16.00 milliliters of a 5.90-molar Pb(NO₃)₂ solution.
c) 508 mL of a 2.75-molar NaF solution.
d) 6.20 L of a 3.76-molar Na₂O solution.
e) 0.500 L of a 1.00 M KCl solution.
f) 4.35 L of a 3.50 M CaCl₂ solution.

Solution set-ups:

a) (0.00300 mol L¯¹) (0.289 L) = x / 187.56 g mol¯¹
b) (5.90 mol L¯¹) (0.01600 L) = x / 331.2 g mol¯¹
c) (2.75 mol L¯¹) (0.508 L) = x / 41.99 g mol¯¹
d) (3.76 mol L¯¹) (6.20 L) = x / 61.98 g mol¯¹
e) (1.00 mol L¯¹) (0.500 L) = x / 74.55 g mol¯¹
f) (3.50 mol L¯¹) (4.35 L) = x / 110.99 g mol¯¹

Problem #19: Determine the final volume of these solutions:

a) 4.67 moles of Li₂SO₃ dissolved to make a 3.89 M solution.
b) 4.907 moles of Al₂O₃ to make a 0.500 M solution.
c) 0.783 grams of Na₂CO₃ to make a 0.348 M solution.
d) 8.97 grams of (NH₄)₂CO₃ to make a 0.250-molar solution.
e) 48.00 grams of PbCl₂ to form a 5.0-molar solution.

Solution set-ups:

a) x = 4.67 mol / 3.89 mol L¯¹
b) x = 4.907 mol / 0.500 mol L¯¹
c) (0.348 mol L¯¹) (x) = 0.783 g / 105.99 g mol¯¹
d) (0.250 mol L¯¹) (x) = 8.97 g / 96.01 g mol¯¹
e) (5.00 mol L¯¹) (x) = 48.0 g / 278.1 g mol¯¹

Problem #20: A student placed 11.0 g of glucose (C₆H₁₂O₆) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500L. How many grams of glucose are in 100. mL of the final solution?

Solution path #1: Netspot wi fi reporter 2 1 472.

1) Calculate molarity of first solution (produced by dissolving 11.0 g of glucose):

MV = grams / molar mass

(x) (0.100 L) = 11.0 g / 180.155 g/mol

x = 0.610585 mol/L (I'll carry a few guard digits.)

2) Calculate molarity of second solution (produced by diluting the first solution):

M₁V₁ = M₂V₂

(0.0200 L) (0.610585 mol/L) = (0.500 L) (x)

x = 0.0244234 mol/L

3) Determine grams of glucose in 100. mL of second solution:

MV = grams / molar mass

(0.0244234 mol/L) (0.100 L) = x / 180.155 g/mol

x = 0.44 g

Solution path #2:

1) Calculate how much glucose you have in 20.0 mL of the first solution.

11.0 g is to 100. mL as x is to 20.0 mL

Cross-multiply and divide

100x = 11.0 times 20.0

x = 2.2 g

2) When you dilute the 20.0 mL sample to 500.0 mL, you have 2.2 g glucose in the solution.2.2 g is to 500. mL as x is to 100. mL

Cross-multiply and divide

500x = 2.2 times 100

x = 0.44 g

In 100 mL of the final solution are 0.44 g glucose.

Problem #21: Commercial bleach solution contains 5.25% (by mass) of NaClO in water. It has a density of 1.08 g/mL. Caculate the molarity of this solution. (Hints: assume you have 1.00 L of solution; molar mass of NaClO 74.4 g/mol)

Solution:

1) Determine mass of 1.00 L of solution:

(1.08 g/mL) (1000 mL) = 1080 g

2) Determine mass of NaClO in 1080 g of solution:

(1080 g) (0.0525) = 56.7 g

3) Determine moles of NaClO:

56.7 g / 74.4 g/mol = 0.762 mol

4) Determine molarity of solution:

0.762 mol / 1.00 L = 0.762 M

Problem #22: What is the molality (and molarity) of a 20.0% by mass hydrochloric acid solution? The density of the solution is 1.0980 g/mL.

Solution:1) Determine moles of HCl in 100.0 g of 20.0% solution.

20.0 % by mass means 20.0 g of HCl in 100.0 g of solution.

20.0 g / 36.4609 g/mol = 0.548 mol

2) Determine molality:

0.548 mol / 0.100 kg = 5.48 m

3) Determine volume of 100.0 g of solution.

100.0 g / 1.0980 g/mL = 91.07468 mL

4) Determine molarity:

0.548 mol / 0.09107468 L = 6.02 m

Problem #23: 25.0 mL of 0.250 M KI, 25.0 mL of 0.100 K₂SO₄, and 15.0 mL of 0.100 M MgCl₂ were mixed together in a beaker. What are the molar concentrations of I¯, Cl¯ and K⁺ in the beaker?

Bartender 3 0 64 Grams =

Solution:

1) Calculate the total volume of the mixed solutions:

25.0 mL + 25.0 mL + 15.0 mL = 65.0 mL

2) Concentration of iodide ion:

M₁V₁ = M₂V₂

(0.250 mol/L) (25.0 mL) = (x) (65.0 mL)

x = 0.09615 M

to three sig figs, 0.0962 M

3) Concentration of the chloride ion:

moles Cl¯ ---> (0.100 mol/L) (0.0150 L) (2 Cl¯ / 1 MgCl₂) = 0.00300 mol

0.00300 mol / 0.065 L = 0.04615 M

to three sig figs, 0.0462 M

4) Concentration of the potassium ion:

moles K⁺ from KI ---> (0.250 mol/L) (0.0250 L) = 0.00625 mol
moles K⁺ from K₂SO₄ ---> (0.100 mol/L) (0.0250 L) (2 K⁺ / 1 K₂SO₄) = 0.00500 mol

0.00625 mol + 0.00500 mol = 0.01125 mol

0.01125 mol / 0.0650 L = 0.173 M

Problem #24: Calculate the total concentration of all the ions in each of the following solutions:

a. 3.25 M NaCl
b. 1.75 M Ca(BrO₃)₂
c. 12.1 g of (NH₄)₂SO₃ in 615 mL in solution.

Solution:

1) the sodium chloride solution:

for every one NaCl that dissolves, two ions are produced (one Na⁺ and one Cl¯).

the total concentration of all ions is this:

(3.25 mol/L) times (2 total ions / 1 NaCl formula unit) = 6.50 M

2) the calcium bromate solution:

three total ions are produced for every one Ca(BrO₃)₂ that dissolves (one Ca²⁺ and two BrO₃¯

the total concentration of all ions is this:

Bartender 3 0 64 Grams Equal

Autodesk flare 2018 2 – advanced 3d visual effects software. (1.75 mol/L) times (3 total ions / 1 Ca(BrO₃)₂ formula unit) = 5.25 M

3) the ammonium sulfite solution:

calculate the concentration of (NH₄)₂SO₃:

(x) (0.615 L) = 12.1 g / 116.1392 g/mol

x = 0.169407 M <--- I'll carry some guard digits

calculate the concentration of all ions:

(NH₄)₂SO₃ produces three ions for every one formula unit that dissolves.

0.169407 M times 3 = 0.508221 M

to three sig figs, 0.508 M

Problem #25: A solution of calcium bromide contains 20.0 g dm^-3. What is the molarity of the solution with respect to calcium bromide and bromine ions.

Solution:

MV = mass / molar mass

(x) (1.00 L) = 20.0 g / 199.886 g/mol

x = 0.100 M

When CaBr₂ ionizes, two bromide ions are released for every one CaBr2 that dissolves. That leads to this:

[Br^-] = 0.200 M

Bartender 3 0 64 Grams