Question

question 5
1 pts
how many grams of calcium hydroxide would you need to prepare a 100.00 ml solution
of 0.575 m ca(oh)2? report your answer in grams to 3 sig figs, but do not include units in the answer.

question 6
1 pts
you wish to prepare 0.849 m kmno4 from a stock solution of potassium permanganate that is 3.382 m.
how many milliliters of the stock solution do you need to make 1.00 l of 0.849 m kmno4? report the
volume to 3 sig figs and do not include units in your answer.

Explanation:

Step1: Calculate moles of $Ca(OH)_2$

Use formula $n = C\times V$. Given $C = 0.575\ M$ and $V=100.00\ mL = 0.10000\ L$.
$n = 0.575\ M\times0.10000\ L=0.0575\ mol$

Step2: Calculate molar mass of $Ca(OH)_2$

Molar mass of $Ca = 40.08\ g/mol$, $O = 16.00\ g/mol$, $H = 1.008\ g/mol$.
$M = 40.08+(2\times16.00)+(2\times1.008)=74.096\ g/mol$

Step3: Calculate mass of $Ca(OH)_2$

Use formula $m = n\times M$.
$m = 0.0575\ mol\times74.096\ g/mol = 4.26\ g$

Step1: Use dilution formula $C_1V_1 = C_2V_2$

Given $C_1 = 3.382\ M$, $C_2 = 0.849\ M$, $V_2 = 1.00\ L$.
Rearrange for $V_1$: $V_1=\frac{C_2V_2}{C_1}$

Step2: Calculate $V_1$

$V_1=\frac{0.849\ M\times1.00\ L}{3.382\ M}=0.251\ L = 251\ mL$

Answer:

4.26