Question

1.) Lactated Ringer's solution (NaCl, KCl, CaCl2, and NaHCO3 in water) for intravenous fluid replacement contains 109mEq/L of Cl-. If the patient received 1250 mL of Ringers solution, how many moles of Cl- ion were given?

2.) How many milliliters of 10% (m/v) NaOH solution would contain 2.2g of NaOH?

3.) How many moles of BaCl2 are in 600 mL of a 2.5 M of BaCl2 solution?

4.) The molarity of a solution of 29.2g of NaCl in 134 mL of solution is ____mol/L.

5.) How many grams of Na2CO3 would be needed to prepare 400mL of 3.00M solution of Na2CO3?

6.) What is the molarity of NaNO3 solution made by diluting 33.0 ml of a 5.50 M solution to a final volume of 250. mL?

7.) A patient receives 100. mL of 20.0% (m/v) mannitol solution every hour. How many grams of mannitol are given in one hour?

8.) One needs a 25.00 mL of a 3.000 ppm Fe solution. How many microliters (mL) of a 4000.0 ppm Fe stock solution are needed to make a the desired solution?

Answer 1

1. To calculate the number of moles of Cl- ion given, we need to first calculate the total number of Cl- ions in 1250 mL of Ringer's solution:

1250 mL x (109 mEq/L / 1000 mL/L) = 136.25 mEq of Cl-

Next, we can convert mEq to moles using the molecular weight of Cl-:

136.25 mEq x (1 mol / 1000 mEq) = 0.13625 mol of Cl-

Therefore, 0.13625 moles of Cl- ion were given.

2. To calculate the volume of 10% NaOH solution containing 2.2 g of NaOH, we need to use the definition of percent concentration:

10 g NaOH / 100 mL solution = 2.2 g NaOH / x mL solution

Solving for x, we get:

x = (2.2 g NaOH) / (10 g NaOH / 100 mL) = 22 mL

Therefore, 22 mL of 10% NaOH solution would contain 2.2 g of NaOH.

3. To calculate the number of moles of BaCl2 in 600 mL of a 2.5 M solution, we can use the definition of molarity:

2.5 mol/L = x mol / 0.6 L

Solving for x, we get:

x = 1.5 mol of BaCl2

Therefore, 1.5 moles of BaCl2 are in 600 mL of a 2.5 M solution.

4. To calculate the molarity of a solution of 29.2 g of NaCl in 134 mL of solution, we need to first convert the mass of NaCl to moles:

29.2 g NaCl x (1 mol / 58.44 g) = 0.499 mol NaCl

Next, we can calculate the volume of the solution in liters:

134 mL x (1 L / 1000 mL) = 0.134 L

Finally, we can calculate the molarity:

0.499 mol / 0.134 L = 3.72 M

Therefore, the molarity of the solution is 3.72 M.

5. To calculate the mass of Na2CO3 needed to prepare a 400 mL of 3.00 M solution of Na2CO3, we can use the molarity formula:

Molarity = moles of solute / liters of solution

Rearranging the formula to solve for moles of solute:

moles of solute = Molarity x liters of solution

moles of Na2CO3 = 3.00 M x 0.400 L = 1.20 moles of Na2CO3

Next, we can convert moles of Na2CO3 to grams using the molar mass of Na2CO3:

1.20 moles Na2CO3 x (105.99 g / 1 mole) = 127.2 g of Na2CO3

Therefore, 127.2 grams of Na2CO3 are needed to prepare a 400 mL of 3.00 M solution of Na2CO3.

6. To calculate the molarity of NaNO3 solution made by diluting 33.0 ml of a 5.50 M solution to a final volume of 250. mL, we can use the dilution formula:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Rearranging the formula to solve for M2:

M2 = M1V1 / V2

M2 = (5.50 M) x (33.0 mL) / (250. mL)

M2 = 0.726 M

Therefore, the molarity of NaNO3 solution made by diluting 33.0 ml of a 5.50 M solution to a final volume of 250. mL is 0.726 M.

7. To calculate the grams of mannitol given in one hour, we can use the definition of percent concentration:

20 g mannitol / 100 mL solution = x g mannitol / 100 mL solution

Simplifying the equation, we get:

x = 20 g mannitol

Therefore, 20 g of mannitol are given in 100 mL of solution. Since the patient receives 100 mL of solution every hour, the amount of mannitol given in one hour is also 20 g.

8. To calculate the microliters of a 4000.0 ppm Fe stock solution