Math / Algebra

QuestionFor an experiment you need to prepare 407 mL of a working solution that contains 622 mM NaCl (molecular mass 58.4 g/mol), 0.85 % w/v glucose (molecular mass 180 g/mol) and 100 µg/mL Penicillin (stock solution 100 mg/mL). You therefore need: Answer: Amount of NaCl = ___ g Answer: Amount of glucose = _ g Answer: Amount of Penicillin = _ μL
How many mmoles of NaCl does this working solution contain? Answer: Number of millimoles of NaCl in the working solution = _ mmoles
What is the molar concentration of glucose in this working solution? Answer: Molar concentration of glucose = _ mM
In the experiment, you add 40 μL of the working solution to exactly 360 μL of water. What is the concentration of NaCl and glucose, respectively, in this solution? Answer: Concentration of NaCl = _ mM Answer: Concentration of Glucose = _____ mM

Studdy Solution

STEP 1

What is this asking? We need to figure out how much of each ingredient (NaCl, glucose, and Penicillin) we need to make a special solution, and then calculate the concentrations of some of the ingredients after we dilute the solution. Watch out! Units are super important here!
We've got grams, moles, milliliters, microliters, percentages... a whole bunch of stuff to keep track of.
Don't mix them up!

STEP 2

1. Calculate NaCl
2. Calculate Glucose
3. Calculate Penicillin
4. Calculate mmoles of NaCl
5. Calculate molar concentration of glucose
6. Calculate diluted NaCl concentration
7. Calculate diluted glucose concentration

STEP 3

We need $\text{\textbf{407 mL}}$ of a solution that's $\text{\textbf{622 mM}}$ NaCl.
Remember, mM means millimolar, which is millimoles per liter.

STEP 4

Let's convert our volume to liters: $\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{1 L}}}{\text{\textbf{1000 mL}}} = \text{\textbf{0.407 L}}$ .

STEP 5

Now, let's find out how many millimoles of NaCl we need: $\text{\textbf{0.407 L}} \cdot \frac{\text{\textbf{622 mmol}}}{\text{\textbf{1 L}}} = \text{\textbf{253.154 mmol}}$ of NaCl.

STEP 6

Finally, let's convert millimoles to grams using the molecular mass of NaCl ( $\text{\textbf{58.4 g/mol}}$ ): $\text{\textbf{253.154 mmol}} \cdot \frac{\text{\textbf{1 mol}}}{\text{\textbf{1000 mmol}}} \cdot \frac{\text{\textbf{58.4 g}}}{\text{\textbf{1 mol}}} = \text{\textbf{14.78 g}}$ of NaCl.

STEP 7

We need a \text{\textbf{0.85% w/v}} glucose solution.
This means $\text{\textbf{0.85 g}}$ of glucose for every $\text{\textbf{100 mL}}$ of solution.

STEP 8

Since we're making $\text{\textbf{407 mL}}$ , we need: $\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{0.85 g}}}{\text{\textbf{100 mL}}} = \text{\textbf{3.46 g}}$ of glucose.

STEP 9

We need $\text{\textbf{100 µg}}$ of Penicillin per $\text{\textbf{1 mL}}$ of solution, and we're making $\text{\textbf{407 mL}}$ .

STEP 10

So, we need a total of $\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{100 µg}}}{\text{\textbf{1 mL}}} = \text{\textbf{40700 µg}}$ of Penicillin.

STEP 11

Our stock solution is $\text{\textbf{100 mg/mL}}$ , which is the same as $\text{\textbf{100,000 µg/mL}}$ (since 1 mg = 1000 µg).

STEP 12

Therefore, we need $\text{\textbf{40700 µg}} \cdot \frac{\text{\textbf{1 mL}}}{\text{\textbf{100000 µg}}} = \text{\textbf{0.407 mL}}$ or $\text{\textbf{407 µL}}$ of the Penicillin stock solution.

STEP 13

We already calculated that we need $\text{\textbf{253.154 mmol}}$ of NaCl back in step one!

STEP 14

We calculated we need $\text{\textbf{3.46 g}}$ of glucose.
Let's convert that to moles using the molecular mass of glucose ( $\text{\textbf{180 g/mol}}$ ): $\text{\textbf{3.46 g}} \cdot \frac{\text{\textbf{1 mol}}}{\text{\textbf{180 g}}} = \text{\textbf{0.0192 mol}}$ .
This is the same as $\text{\textbf{19.2 mmol}}$ .

STEP 15

Since our solution is $\text{\textbf{407 mL}}$ or $\text{\textbf{0.407 L}}$ , the molar concentration is: $\frac{\text{\textbf{19.2 mmol}}}{\text{\textbf{0.407 L}}} = \text{\textbf{47.2 mM}}$ .

STEP 16

We're adding $\text{\textbf{40 µL}}$ of our working solution to $\text{\textbf{360 µL}}$ of water, for a total volume of $\text{\textbf{400 µL}}$ .

STEP 17

The dilution factor is $\frac{\text{\textbf{40 µL}}}{\text{\textbf{400 µL}}} = \text{\textbf{0.1}}$ .

STEP 18

So, the new NaCl concentration is $\text{\textbf{622 mM}} \cdot \text{\textbf{0.1}} = \text{\textbf{62.2 mM}}$ .

STEP 19

Using the same dilution factor of $\text{\textbf{0.1}}$ from the previous step, the new glucose concentration is $\text{\textbf{47.2 mM}} \cdot \text{\textbf{0.1}} = \text{\textbf{4.72 mM}}$ .

STEP 20

Amount of NaCl = $\text{\textbf{14.78 g}}$ Amount of glucose = $\text{\textbf{3.46 g}}$ Amount of Penicillin = $\text{\textbf{407 μL}}$ Number of millimoles of NaCl in the working solution = $\text{\textbf{253.154 mmoles}}$ Molar concentration of glucose = $\text{\textbf{47.2 mM}}$ Concentration of NaCl = $\text{\textbf{62.2 mM}}$ Concentration of Glucose = $\text{\textbf{4.72 mM}}$

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Studdy Solution

STEP 1

STEP 2

1. Calculate NaCl2. Calculate Glucose3. Calculate Penicillin4. Calculate mmoles of NaCl5. Calculate molar concentration of glucose6. Calculate diluted NaCl concentration7. Calculate diluted glucose concentration

STEP 3

We need 407 mL\text{\textbf{407 mL}}407 mL of a solution that's 622 mM\text{\textbf{622 mM}}622 mM NaCl.Remember, mM means millimolar, which is millimoles per liter.

STEP 4

Let's convert our volume to liters: 407 mL⋅1 L1000 mL=0.407 L\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{1 L}}}{\text{\textbf{1000 mL}}} = \text{\textbf{0.407 L}}407 mL⋅1000 mL1 L​=0.407 L.

STEP 5

Now, let's find out how many millimoles of NaCl we need: 0.407 L⋅622 mmol1 L=253.154 mmol\text{\textbf{0.407 L}} \cdot \frac{\text{\textbf{622 mmol}}}{\text{\textbf{1 L}}} = \text{\textbf{253.154 mmol}}0.407 L⋅1 L622 mmol​=253.154 mmol of NaCl.

STEP 6

STEP 7

We need a \text{\textbf{0.85% w/v}} glucose solution.This means 0.85 g\text{\textbf{0.85 g}}0.85 g of glucose for every 100 mL\text{\textbf{100 mL}}100 mL of solution.

STEP 8

Since we're making 407 mL\text{\textbf{407 mL}}407 mL, we need: 407 mL⋅0.85 g100 mL=3.46 g\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{0.85 g}}}{\text{\textbf{100 mL}}} = \text{\textbf{3.46 g}}407 mL⋅100 mL0.85 g​=3.46 g of glucose.

STEP 9

We need 100 µg\text{\textbf{100 µg}}100 µg of Penicillin per 1 mL\text{\textbf{1 mL}}1 mL of solution, and we're making 407 mL\text{\textbf{407 mL}}407 mL.

STEP 10

So, we need a total of 407 mL⋅100 µg1 mL=40700 µg\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{100 µg}}}{\text{\textbf{1 mL}}} = \text{\textbf{40700 µg}}407 mL⋅1 mL100 µg​=40700 µg of Penicillin.

STEP 11

Our stock solution is 100 mg/mL\text{\textbf{100 mg/mL}}100 mg/mL, which is the same as 100,000 µg/mL\text{\textbf{100,000 µg/mL}}100,000 µg/mL (since 1 mg = 1000 µg).

STEP 12

Therefore, we need 40700 µg⋅1 mL100000 µg=0.407 mL\text{\textbf{40700 µg}} \cdot \frac{\text{\textbf{1 mL}}}{\text{\textbf{100000 µg}}} = \text{\textbf{0.407 mL}}40700 µg⋅100000 µg1 mL​=0.407 mL or 407 µL\text{\textbf{407 µL}}407 µL of the Penicillin stock solution.

STEP 13

We already calculated that we need 253.154 mmol\text{\textbf{253.154 mmol}}253.154 mmol of NaCl back in step one!

STEP 14

STEP 15

Since our solution is 407 mL\text{\textbf{407 mL}}407 mL or 0.407 L\text{\textbf{0.407 L}}0.407 L, the molar concentration is: 19.2 mmol0.407 L=47.2 mM\frac{\text{\textbf{19.2 mmol}}}{\text{\textbf{0.407 L}}} = \text{\textbf{47.2 mM}}0.407 L19.2 mmol​=47.2 mM.

STEP 16

We're adding 40 µL\text{\textbf{40 µL}}40 µL of our working solution to 360 µL\text{\textbf{360 µL}}360 µL of water, for a total volume of 400 µL\text{\textbf{400 µL}}400 µL.

STEP 17

The dilution factor is 40 µL400 µL=0.1\frac{\text{\textbf{40 µL}}}{\text{\textbf{400 µL}}} = \text{\textbf{0.1}}400 µL40 µL​=0.1.

STEP 18

So, the new NaCl concentration is 622 mM⋅0.1=62.2 mM\text{\textbf{622 mM}} \cdot \text{\textbf{0.1}} = \text{\textbf{62.2 mM}}622 mM⋅0.1=62.2 mM.

STEP 19

Using the same dilution factor of 0.1\text{\textbf{0.1}}0.1 from the previous step, the new glucose concentration is 47.2 mM⋅0.1=4.72 mM\text{\textbf{47.2 mM}} \cdot \text{\textbf{0.1}} = \text{\textbf{4.72 mM}}47.2 mM⋅0.1=4.72 mM.

STEP 20

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1. Calculate NaCl
2. Calculate Glucose
3. Calculate Penicillin
4. Calculate mmoles of NaCl
5. Calculate molar concentration of glucose
6. Calculate diluted NaCl concentration
7. Calculate diluted glucose concentration

We need $\text{\textbf{407 mL}}$ of a solution that's $\text{\textbf{622 mM}}$ NaCl.
Remember, mM means millimolar, which is millimoles per liter.

Let's convert our volume to liters: $\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{1 L}}}{\text{\textbf{1000 mL}}} = \text{\textbf{0.407 L}}$ .

Now, let's find out how many millimoles of NaCl we need: $\text{\textbf{0.407 L}} \cdot \frac{\text{\textbf{622 mmol}}}{\text{\textbf{1 L}}} = \text{\textbf{253.154 mmol}}$ of NaCl.

We need a \text{\textbf{0.85% w/v}} glucose solution.
This means $\text{\textbf{0.85 g}}$ of glucose for every $\text{\textbf{100 mL}}$ of solution.

Since we're making $\text{\textbf{407 mL}}$ , we need: $\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{0.85 g}}}{\text{\textbf{100 mL}}} = \text{\textbf{3.46 g}}$ of glucose.

We need $\text{\textbf{100 µg}}$ of Penicillin per $\text{\textbf{1 mL}}$ of solution, and we're making $\text{\textbf{407 mL}}$ .

So, we need a total of $\text{\textbf{407 mL}} \cdot \frac{\text{\textbf{100 µg}}}{\text{\textbf{1 mL}}} = \text{\textbf{40700 µg}}$ of Penicillin.

Our stock solution is $\text{\textbf{100 mg/mL}}$ , which is the same as $\text{\textbf{100,000 µg/mL}}$ (since 1 mg = 1000 µg).

Therefore, we need $\text{\textbf{40700 µg}} \cdot \frac{\text{\textbf{1 mL}}}{\text{\textbf{100000 µg}}} = \text{\textbf{0.407 mL}}$ or $\text{\textbf{407 µL}}$ of the Penicillin stock solution.

We already calculated that we need $\text{\textbf{253.154 mmol}}$ of NaCl back in step one!

Since our solution is $\text{\textbf{407 mL}}$ or $\text{\textbf{0.407 L}}$ , the molar concentration is: $\frac{\text{\textbf{19.2 mmol}}}{\text{\textbf{0.407 L}}} = \text{\textbf{47.2 mM}}$ .

We're adding $\text{\textbf{40 µL}}$ of our working solution to $\text{\textbf{360 µL}}$ of water, for a total volume of $\text{\textbf{400 µL}}$ .

The dilution factor is $\frac{\text{\textbf{40 µL}}}{\text{\textbf{400 µL}}} = \text{\textbf{0.1}}$ .

So, the new NaCl concentration is $\text{\textbf{622 mM}} \cdot \text{\textbf{0.1}} = \text{\textbf{62.2 mM}}$ .

Using the same dilution factor of $\text{\textbf{0.1}}$ from the previous step, the new glucose concentration is $\text{\textbf{47.2 mM}} \cdot \text{\textbf{0.1}} = \text{\textbf{4.72 mM}}$ .