8. what mass of ammonium phosphate is present in 1.80 l of a 145 - mmol/l solution?\n9. what volume is…
8. what mass of ammonium phosphate is present in 1.80 l of a 145 - mmol/l solution?\n9. what volume is required to make a 0.48 - mol/l solution with 24 g of sodium hydroxide?\n10. determine the amount concentration of the iron ions and sulfate ions in a 0.62 - mol/l solution of iron(iii) sulfate.\n11. 18 ml of a 0.055 - mol/l sulfuric acid solution is diluted up to 30 ml. what is the new concentration of the acid?
Answer
8.
Explanation:
Step1: Calculate moles of ammonium phosphate
We know that $n = c\times V$, where $c$ is concentration and $V$ is volume. Given $c = 145\ mmol/L=0.145\ mol/L$ and $V = 1.80\ L$. So $n=0.145\ mol/L\times1.80\ L = 0.261\ mol$.
Step2: Find molar - mass of ammonium phosphate
The formula of ammonium phosphate is $(NH_{4}){3}PO{4}$. The molar - mass $M=(3\times(14 + 4\times1))+31+(4\times16)=149\ g/mol$.
Step3: Calculate mass of ammonium phosphate
Using $m = n\times M$, we have $m = 0.261\ mol\times149\ g/mol=38.889\ g\approx38.9\ g$.
Answer:
$38.9\ g$
9.
Explanation:
Step1: Calculate moles of sodium hydroxide
The molar - mass of $NaOH$ is $M = 23+16 + 1=40\ g/mol$. Given $m = 24\ g$, then $n=\frac{m}{M}=\frac{24\ g}{40\ g/mol}=0.6\ mol$.
Step2: Calculate volume of the solution
We know that $V=\frac{n}{c}$, where $c = 0.48\ mol/L$ and $n = 0.6\ mol$. So $V=\frac{0.6\ mol}{0.48\ mol/L}=1.25\ L$.
Answer:
$1.25\ L$
10.
Explanation:
Step1: Write the dissociation equation of iron(III) sulfate
The formula of iron(III) sulfate is $Fe_{2}(SO_{4}){3}$, and its dissociation equation in water is $Fe{2}(SO_{4}){3}\rightarrow2Fe^{3 + }+3SO{4}^{2 - }$.
Step2: Calculate concentration of iron ions
From the dissociation equation, the mole - ratio of $Fe_{2}(SO_{4}){3}$ to $Fe^{3+}$ is $1:2$. Given $c(Fe{2}(SO_{4})_{3}) = 0.62\ mol/L$, then $c(Fe^{3+})=2\times0.62\ mol/L = 1.24\ mol/L$.
Step3: Calculate concentration of sulfate ions
The mole - ratio of $Fe_{2}(SO_{4}){3}$ to $SO{4}^{2 - }$ is $1:3$. So $c(SO_{4}^{2 - })=3\times0.62\ mol/L = 1.86\ mol/L$.
Answer:
$c(Fe^{3+}) = 1.24\ mol/L$, $c(SO_{4}^{2 - })=1.86\ mol/L$
11.
Explanation:
Step1: Use the dilution formula
The dilution formula is $c_1V_1=c_2V_2$, where $c_1 = 0.055\ mol/L$, $V_1 = 18\ mL = 0.018\ L$, and $V_2 = 30\ mL=0.030\ L$.
Step2: Solve for $c_2$
We can re - arrange the dilution formula to $c_2=\frac{c_1V_1}{V_2}$. Substituting the values, we get $c_2=\frac{0.055\ mol/L\times0.018\ L}{0.030\ L}=0.033\ mol/L$.
Answer:
$0.033\ mol/L$