14. list the steps needed to make a 150 ml solution of cacl₂ with a molarity of 2m.\n15. what volume of a 12…

14.

Explanation:

Step1: Calculate moles of CaCl₂

Use the formula $n = M\times V$. Given $M = 2M$ and $V=150\ mL = 0.15\ L$. So $n = 2\ mol/L\times0.15\ L=0.3\ mol$.

Step2: Calculate mass of CaCl₂

The molar - mass of $CaCl_{2}$ is $M_{CaCl_{2}}=40.08\ g/mol + 2\times35.45\ g/mol=110.98\ g/mol$. Mass $m=n\times M_{CaCl_{2}}$, so $m = 0.3\ mol\times110.98\ g/mol = 33.294\ g$.

Step3: Prepare the solution

1. Weigh out $33.294\ g$ of $CaCl_{2}$ using a balance.
2. Transfer the weighed $CaCl_{2}$ to a $150 - mL$ volumetric flask.
3. Add some distilled water to dissolve the $CaCl_{2}$ completely.
4. Then add more distilled water until the volume of the solution reaches the $150 - mL$ mark on the volumetric flask. Mix well.

15.

Explanation:

Step1: Use the dilution formula $M_1V_1 = M_2V_2$

Here, $M_1 = 12\ M$, $M_2 = 3.0\ M$, and $V_2=250\ mL$. We need to find $V_1$.

Step2: Rearrange the formula to solve for $V_1$

$V_1=\frac{M_2V_2}{M_1}$. Substitute the values: $V_1=\frac{3.0\ M\times250\ mL}{12\ M}=62.5\ mL$.

Answer:

$62.5\ mL$

16.

Explanation:

Step1: Analyze the dissociation of $CaCl_{2}$

$CaCl_{2}\rightarrow Ca^{2 +}+2Cl^{-}$. For every 1 mole of $CaCl_{2}$, 2 moles of $Cl^{-}$ are produced.

Step2: Calculate the concentration of $Cl^{-}$

Given the concentration of $CaCl_{2}$ is $5\ M$. So the concentration of $Cl^{-}$ is $2\times5\ M = 10\ M$.

Answer:

$10\ M$

17.

Explanation:

Step1: Analyze the dissociation of $Al(NO_{3})_{3}$

$Al(NO_{3}){3}\rightarrow Al^{3+}+3NO{3}^{-}$. For every 1 mole of $Al(NO_{3}){3}$, 3 moles of $NO{3}^{-}$ are produced.

Step2: Calculate the concentration of $NO_{3}^{-}$

Given the concentration of $Al(NO_{3}){3}$ is $1.5\ M$. So the concentration of $NO{3}^{-}$ is $3\times1.5\ M = 4.5\ M$.

Answer:

$4.5\ M$