Question

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

Solution

## 11.1 g $$CaCl_2$$ and 29.25 g of NaCl are diluted with water of 100 mL$$[Ca^{2+}]=0.001M$$$$[Na^{\oplus}=0.005 M$$$$[Cl^{\ominus}=0.007 M$$i. Moles of $$CaCl_2=\frac {11.1}{111}=0.1 mol CaCl_2$$$$=0.1 mol Ca^{2+}+0.2 mol Cl^{\ominus}$$ii. Moles of $$NaCl=\frac {29.25}{58.5}=0.5 mol NaCl$$$$=0.5 mol Na^{\oplus}+0.5 mol Cl^{\ominus}$$Adding (i) and (ii), we get$$0.1 mol Ca^{2+}+0.5 mol Na^{\oplus}+0.7 mol Cl^{\ominus}$$$$\therefore [Ca^{2+}]=\frac {0.1}{100}=0.001 M$$$$[Na^{\oplus}]=\frac {0.5}{100}=0.005 M$$$$[Cl^{\ominus}]=\frac {0.7}{100}=0.007 M$$3.0L of 4.0 M NaCl and 4.0 L of 2.0 M $$CaCl_2$$ are combined and diluted to 10.0L$$[Ca^{2+}]=0.8M$$$$[Na^{\oplus}=1.2 M$$$$[Cl^{\ominus}=2.8M$$ i. $$NaCl\Rightarrow 3\times 4=12\Rightarrow Na^{\oplus}=12 mol, Cl^{\ominus}=12 mol$$ii. $$CaCl_2\Rightarrow 4\times 2=8\Rightarrow Ca^{2+}=8 mol, Cl^{\ominus}=16 mol$$Adding (i) and (ii), we get$$12 mol Na^{\oplus}+8 mol Ca^{2+}+28 mol Cl^{\ominus}$$$$therefore [Ca^{2+}]=\frac {8}{10}=0.8 M$$$$[Na^{\oplus}]=\frac {12}{10}=1.2 M$$$$[Cl^{\ominus}]=\frac {28}{10}=2.8 M$$300 mL of3.0 M NaCl is added to 200 mL of 4.0 M $$CaCl_2$$$$[Ca^{2+}]=1.6 M$$$$[Na^{\oplus}=1.8 M$$$$[Cl^{\ominus}=5.0 M$$i. $$NaCl=300\times 3=900 mmol\Rightarrow 900 mmol$$$$Na^{\oplus}+900 mmol Cl^{\ominus}$$$$CaCl_2=200\times 4=800 mmol\Rightarrow 800 mmol$$$$Ca^{2+}=1600 mmol Cl^{\ominus}$$Adding (i) and (ii), we get$$800 m mol Ca^{2+}+900 mmol Na^{\oplus}+2500 mmol Cl^{\ominus}$$Total volume $$=300+200=500 mL$$$$\therefore [Ca^{2+}]=\frac {800 mmol}{500 mL}=1.6 M$$$$[Na^{\oplus}]=\frac {900 mmol}{500 mL}=1.8 M$$$$[Cl^{\ominus}]=\frac {2500 mmol}{500}=5 M$$100 mL of 2.0 M HCL +200mL of 1.0M NaOH + 150 mL of 4.0 M $$CaCl_2$$ + 50 mL of $$H_2O$$$$[Ca^{2+}]=1.2 M$$$$[Na^{\oplus}=0.4 M$$$$[Cl^{\ominus}=2.8 M$$ i. $$HCl=100\times 2=200 mmol\Rightarrow 200 mmol H^{\oplus}+200 mmol Cl^{\ominus}$$ii. $$NaOH=200\times 1=200 mmol\Rightarrow 200 mmol Na^{\oplus}+200 mmol \overset {\ominus}{O}H$$Since $$H^{\oplus}$$ and $$\overset {\ominus}{O}H$$ react to give $$H_2O$$.$$\underset {200 mmol}{H^{\oplus}}+\underset {200 mmol}{\overset {\ominus}{O}H}\Rightarrow \underset {200 mmol}{H_2O}$$$$Cl^{\ominus}=200 mmol, Na^{\oplus}\Rightarrow 200 mmol$$iii. $$CaCl_2\Rightarrow 150\times 4=600=mmol\Rightarrow 600 mmol$$$$Ca^{2+}+1200 mmol Cl^{\ominus}$$iv. Total volume $$=200+100+150+50$$$$=500 mL$$v. $$Total Ca^{2+}=600 mmol$$$$\therefore [Ca^{2+}]=\frac {600}{500}=1.2 M$$Total $$Na^{\oplus}=200 mmol$$$$\therefore Na^{\oplus}=\frac {200}{500}=0.4 M$$Total $$Cl^{\ominus}=200+1200=1400 mmol$$$$\therefore Cl^{\ominus}=\frac {1400}{500}=2.8 M$$Chemistry

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