Question

review math skills:

1. report the number of sigfigs:
2. 2.30×10⁵ 2.3000 650 30,000 320.0 0.0010 8.60×10⁵
3. use the international system of units (si) to convert:

a. 5.5 g = mg f. 2.50 ml = μl k. 350μl = ml
b. 12.6 g = mg g. 0.50 ml = l l. 650μl = ml
c. 4.5 kg = g h. 10.5ml = l m. 850mg = g

1. find the mass to prepare a 3.0% solution of sucrose in 30 ml of di water.
2. calculate the mass of albumin to prepare a 2.5% solution in 15 ml of sterilized di water. report your answer to one decimal point.
3. find the mass in mg to prepare a 5.5 mg/ml of protease in 15 ml of di water.
4. calculate the mass of kanamycin to prepare a 10 mg/ml in 4.5 ml of solution.
5. find the mass in grams of agarose to prepare a 1.0% agarose in 40 ml of 1x tae buffer.
6. dilutions: find the volume to prepare a 1x tae buffer in 500 ml of di water using a stock solution of 50x tae buffer. report your answer to one decimal point.
7. determine the volume in ml to prepare a 1x tgs buffer in 40 ml of di water using a stock solution of 10x tgs buffer. report your answer to one decimal point.
8. prepare a 0.1m solution of hcl in 30 ml of di water using a stock solution of 2.0m hcl

Explanation:

Step1: Determine significant - figures

Count the non - zero digits and zeros between non - zero digits or trailing zeros after a decimal point.

Step2: Unit conversions

Use the conversion factors between grams, milligrams, kilograms, liters, milliliters, and microliters.

Step3: Mass calculations for percentage solutions

Use the percentage concentration formula (or approximation for dilute solutions) to find the mass of the solute.

Step4: Mass calculations for mass/volume solutions

Multiply the concentration by the volume to get the mass.

Step5: Dilution calculations

Use the dilution formula $C_1V_1 = C_2V_2$ to find the volume of the stock solution needed for dilution.

Answer:

1.

• $2.30\times 10^{5}$: 3 sig - figs
• $2.3000$: 5 sig - figs
• $30000$: 1 sig - fig
• $600.$: 3 sig - figs
• $650$: 2 sig - figs
• $320.0$: 4 sig - figs
• $0.0010$: 2 sig - figs
• $8.60\times 10^{5}$: 3 sig - figs

2.

• a. $5.5\ g = 5500\ mg$ (since $1\ g=1000\ mg$, so $5.5\times1000 = 5500$)
• b. $12.6\ g=12600\ mg$ ($12.6\times1000 = 12600$)
• c. $4.5\ Kg = 4500\ g$ (as $1\ Kg = 1000\ g$, $4.5\times1000=4500$)
• f. $2.50\ mL = 2500\ \mu L$ ($1\ mL = 1000\ \mu L$, $2.50\times1000 = 2500$)
• g. $0.50\ mL=0.0005\ L$ ($1\ L = 1000\ mL$, $0.50\div1000=0.0005$)
• h. $10.5\ mL = 0.0105\ L$ ($10.5\div1000 = 0.0105$)
• k. $350\ \mu L=0.35\ mL$ ($1\ mL = 1000\ \mu L$, $350\div1000 = 0.35$)
• l. $650\ \mu L = 0.65\ mL$ ($650\div1000=0.65$)
• m. $850\ mg = 0.85\ g$ ($850\div1000 = 0.85$)

1. Let the mass of sucrose be $m$. A $3.0\%$ solution means $\frac{m}{m + 30}=0.03$. Since the mass of water is much larger than the mass of solute for a dilute - solution approximation, we can assume $\frac{m}{30}=0.03$. So $m=30\times0.03 = 0.9\ g$.
2. Let the mass of albumin be $m$. A $2.5\%$ solution in $15\ mL$ of water. Using the approximation $\frac{m}{15}=0.025$, then $m = 15\times0.025=0.4\ g$ (to one decimal point).
3. If the concentration is $5.5\ mg/mL$ and the volume is $15\ mL$, then the mass $m = 5.5\times15=82.5\ mg$.
4. If the concentration is $10\ mg/mL$ and the volume is $4.5\ mL$, then the mass $m = 10\times4.5 = 45\ mg$.
5. For a $1.0\%$ agarose solution in $40\ mL$ of $1X$ TAE buffer. Using the approximation $\frac{m}{40}=0.01$, so $m = 40\times0.01=0.4\ g$.
6. Using the dilution formula $C_1V_1=C_2V_2$. Here $C_1 = 50X$, $C_2 = 1X$, $V_2=500\ mL$. Then $V_1=\frac{C_2V_2}{C_1}=\frac{1\times500}{50}=10.0\ mL$.
7. Using the dilution formula $C_1V_1=C_2V_2$. Here $C_1 = 10X$, $C_2 = 1X$, $V_2 = 40\ mL$. Then $V_1=\frac{C_2V_2}{C_1}=\frac{1\times40}{10}=4.0\ mL$.
8. Using the dilution formula $C_1V_1=C_2V_2$. Here $C_1 = 2.0\ M$, $C_2 = 0.1\ M$, $V_2 = 30\ mL$. Then $V_1=\frac{C_2V_2}{C_1}=\frac{0.1\times30}{2.0}=1.5\ mL$.