91Ó°ÊÓ

$1$L of a $4$M $HN{O}_3$ solution is added to the water so that the final volume is $8$L.

Using the concentration formula,

$C_1{V}_1=C_2{V}_2$

role="math" localid="1653929520012" $C_1=4\mathrm{M},V_1=1\mathrm{L},{\mathrm{V}}_2=8\mathrm{L}$

The final concentration is,

$c_2=\frac{c_1{v}_1}{v_2}=4\mathrm{M}×\frac{1L}{8L}=0.5\text{M}$

Water is added to $0.25$L of a $6$ NaF solution to make $2$L of a diluted NaF solution.

Using the concentration formula,

$C_1{V}_1=C_2{V}_2$

role="math" localid="1653929779239" $C_1=6M,V_1=0.25L,V_2=2L$

The final concentration is,

$c_2=\frac{c_1{v}_1}{v_2}=6\mathrm{M}×\frac{0.25\mathrm{L}}{2\mathrm{L}}=0.75\mathrm{M}$

A $50$mL sample of an $8$% KBr solution is diluted with water so that the final volume is $200$mL.

Given,

$C_1=8\mathrm{\%}(M/V),V_1=50\mathrm{mL},V_2=200\mathrm{mL}$

Using the dilution factor = $\frac{V_2}{V_1}$

The dilution factor =$\frac{200}{50}=4$

Now, The concentration is

$\mathrm{\%}M/V=\frac{1}{4}×8\mathrm{\%}=2\%$

A $5$mL sample of a $50$% acetic acid solution is added to water to give a final volume of $25$mL.

Given,$C_1=50\mathrm{\%}(M/V),V_1=5\mathrm{mL},V_2=25\mathrm{mL}$

Using the dilution factor = $\frac{V_2}{V_1}$

the dilution factor = role="math" localid="1653930566200" $\frac{25}{5}=5$

Now, The concentration is

$\mathrm{\%}M/V=\frac{1}{5}×50\mathrm{\%}=10\%$