Question

Match the solution mixtures given in list I with the concentrations given in list II.

Answer 1

11.1 g $CaC{l}_2$ and 29.25 g of NaCl are diluted with water of 100 mL
$\left[C{a}^{2+}\right]=0.001M$
$[N{a}^{\oplus }=0.005M$
$[C{l}^{\ominus }=0.007M$
i. Moles of $CaC{l}_2=\frac{11.1}{111}=0.1molCaC{l}_2$
$=0.1molC{a}^{2+}+0.2molC{l}^{\ominus }$
ii. Moles of $NaCl=\frac{29.25}{58.5}=0.5molNaCl$
$=0.5molN{a}^{\oplus }+0.5molC{l}^{\ominus }$
Adding (i) and (ii), we get
$0.1molC{a}^{2+}+0.5molN{a}^{\oplus }+0.7molC{l}^{\ominus }$
$\therefore \left[C{a}^{2+}\right]=\frac{0.1}{100}=0.001M$
$\left[N{a}^{\oplus }\right]=\frac{0.5}{100}=0.005M$
$\left[C{l}^{\ominus }\right]=\frac{0.7}{100}=0.007M$
3.0L of 4.0 M NaCl and 4.0 L of 2.0 M $CaC{l}_2$ are combined and diluted to 10.0L
$\left[C{a}^{2+}\right]=0.8M$
$[N{a}^{\oplus }=1.2M$
$[C{l}^{\ominus }=2.8M$
i. $NaCl\Rightarrow 3\times 4=12\Rightarrow N{a}^{\oplus }=12mol,C{l}^{\ominus }=12mol$
ii. $CaC{l}_2\Rightarrow 4\times 2=8\Rightarrow C{a}^{2+}=8mol,C{l}^{\ominus }=16mol$
Adding (i) and (ii), we get
$12molN{a}^{\oplus }+8molC{a}^{2+}+28molC{l}^{\ominus }$
$therefore\left[C{a}^{2+}\right]=\frac{8}{10}=0.8M$
$\left[N{a}^{\oplus }\right]=\frac{12}{10}=1.2M$
$\left[C{l}^{\ominus }\right]=\frac{28}{10}=2.8M$
300 mL of3.0 M NaCl is added to 200 mL of 4.0 M $CaC{l}_2$
$\left[C{a}^{2+}\right]=1.6M$
$[N{a}^{\oplus }=1.8M$
$[C{l}^{\ominus }=5.0M$
i. $NaCl=300\times 3=900mmol\Rightarrow 900mmol$
$N{a}^{\oplus }+900mmolC{l}^{\ominus }$
$CaC{l}_2=200\times 4=800mmol\Rightarrow 800mmol$
$C{a}^{2+}=1600mmolC{l}^{\ominus }$
Adding (i) and (ii), we get
$800mmolC{a}^{2+}+900mmolN{a}^{\oplus }+2500mmolC{l}^{\ominus }$
Total volume $=300+200=500mL$
$\therefore \left[C{a}^{2+}\right]=\frac{800mmol}{500mL}=1.6M$
$\left[N{a}^{\oplus }\right]=\frac{900mmol}{500mL}=1.8M$
$\left[C{l}^{\ominus }\right]=\frac{2500mmol}{500}=5M$
100 mL of 2.0 M HCL +200mL of 1.0M NaOH + 150 mL of 4.0 M $CaC{l}_2$ + 50 mL of $H_{2}O$
$\left[C{a}^{2+}\right]=1.2M$
$[N{a}^{\oplus }=0.4M$
$[C{l}^{\ominus }=2.8M$
i. $HCl=100\times 2=200mmol\Rightarrow 200mmol{H}^{\oplus }+200mmolC{l}^{\ominus }$
ii. $NaOH=200\times 1=200mmol\Rightarrow 200mmolN{a}^{\oplus }+200mmol\stackrel{\ominus }{O}H$
Since $H^{\oplus }$ and $\stackrel{\ominus }{O}H$ react to give $H_{2}O$.
$\underset{200mmol}{H^{\oplus }}+\underset{200mmol}{\stackrel{\ominus }{O}H}\Rightarrow \underset{200mmol}{H_{2}O}$
$C{l}^{\ominus }=200mmol,N{a}^{\oplus }\Rightarrow 200mmol$
iii. $CaC{l}_2\Rightarrow 150\times 4=600=mmol\Rightarrow 600mmol$
$C{a}^{2+}+1200mmolC{l}^{\ominus }$
iv. Total volume $=200+100+150+50$
$=500mL$
v. $TotalC{a}^{2+}=600mmol$
$\therefore \left[C{a}^{2+}\right]=\frac{600}{500}=1.2M$
Total $N{a}^{\oplus }=200mmol$
$\therefore N{a}^{\oplus }=\frac{200}{500}=0.4M$
Total $C{l}^{\ominus }=200+1200=1400mmol$
$\therefore C{l}^{\ominus }=\frac{1400}{500}=2.8M$