Question

1. find the mass to prepare a 1m solution of kcl (potassium chloride) in 10 ml of di water. (molar mass of kcl = 74.55 g/mol)
2. convert:

a. 3.0 ml = ________ μl 450 μl = ________ ml
b. 2.5 g = ________ mg 3.5 mg = ________ μg (micro grams)

Explanation:

:

1. 0.7455 g

9.
a. 3000 μL; 0.45 mL
b. 2500 mg; 3500 μg

Step1: Calculate moles of KCl in 8

Use molarity formula $n = M\times V$. Given $M = 1M$, $V=10\ mL=0.01\ L$. So $n = 1\ mol/L\times0.01\ L = 0.01\ mol$.

Step2: Calculate mass of KCl in 8

Use $m=n\times M_m$. Given $n = 0.01\ mol$ and $M_m = 74.55\ g/mol$. So $m=0.01\ mol\times74.55\ g/mol = 0.7455\ g$.

Step3: Convert mL to μL in 9a

Use conversion factor $1\ mL = 1000\ μL$. For $3.0\ mL$, $3.0\ mL\times1000\ μL/mL=3000\ μL$.

Step4: Convert μL to mL in 9a

Use conversion factor $1\ μL=\frac{1}{1000}\ mL$. For $450\ μL$, $450\ μL\times\frac{1}{1000}\ mL/μL = 0.45\ mL$.

Step5: Convert g to mg in 9b

Use conversion factor $1\ g = 1000\ mg$. For $2.5\ g$, $2.5\ g\times1000\ mg/g = 2500\ mg$.

Step6: Convert mg to μg in 9b

Use conversion factor $1\ mg = 1000\ μg$. For $3.5\ mg$, $3.5\ mg\times1000\ μg/mg=3500\ μg$.

Answer:

Step1: Calculate moles of KCl in 8

Use molarity formula $n = M\times V$. Given $M = 1M$, $V=10\ mL=0.01\ L$. So $n = 1\ mol/L\times0.01\ L = 0.01\ mol$.

Step2: Calculate mass of KCl in 8

Use $m=n\times M_m$. Given $n = 0.01\ mol$ and $M_m = 74.55\ g/mol$. So $m=0.01\ mol\times74.55\ g/mol = 0.7455\ g$.

Step3: Convert mL to μL in 9a

Use conversion factor $1\ mL = 1000\ μL$. For $3.0\ mL$, $3.0\ mL\times1000\ μL/mL=3000\ μL$.

Step4: Convert μL to mL in 9a

Use conversion factor $1\ μL=\frac{1}{1000}\ mL$. For $450\ μL$, $450\ μL\times\frac{1}{1000}\ mL/μL = 0.45\ mL$.

Step5: Convert g to mg in 9b

Use conversion factor $1\ g = 1000\ mg$. For $2.5\ g$, $2.5\ g\times1000\ mg/g = 2500\ mg$.

Step6: Convert mg to μg in 9b

Use conversion factor $1\ mg = 1000\ μg$. For $3.5\ mg$, $3.5\ mg\times1000\ μg/mg=3500\ μg$.