Chemistry - Concentration Terms Question with Solution | TestHub

(I) $2\mathrm{M}$ aq. NaOH solution $(\mathrm{d}=1.2\mathrm{gm}/\mathrm{ml})$

2 moles NaOH present in $1l(1000\mathrm{ml})$ solution

or 2 moles $\mathrm{NaOH}(80\mathrm{gm})$ in 1200 gm solution

$1000\mathrm{m}/$ solution $⟶80\mathrm{gm}\mathrm{NaOH}$

100 ml solution $⟶8\mathrm{gm}\mathrm{NaOH}$

1200 gm solution $⟶80\mathrm{gm}$

240 gm solution $\to \frac{80}{1200}\times 240=16\mathrm{gm}\mathrm{NaOH}$

(II) $8\%\frac{\mathrm{W}}{\mathrm{V}}$ aq. $\mathrm{KOH}(\mathrm{d}=1.2\mathrm{gm}/\mathrm{mol})$

8 gm KOH in 100 ml solution

8 gm KOH in 120 gm solution

16 gm KOH in 240 gm solution

(III) $25\%\left(\frac{\mathrm{W}}{\mathrm{W}}\right){\mathrm{CaCO}}_3$ solution $(\mathrm{d}=1.2\mathrm{gm}/\mathrm{ml})$

100 gm solution contains 25 gm CaCO₃

400 gm solution $\to 100\mathrm{gm}$ ( 1 mole) ${\mathrm{CaCO}}_3$

240 gm solution $\to \frac{25}{100}\times 240=60\mathrm{gm}{\mathrm{CaCO}}_3$

$\frac{100}{1.2}\mathrm{ml}$ solution $\to 25\mathrm{gm}{\mathrm{CaCO}}_3$

(IV) ${\mathrm{X}}_{{\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}}=\frac{1}{11}=\frac{1}{1+10}$ :

$\begin{array}{llc}\text{ i.e }& 1\text{ mole }{\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}+10\text{ mole }{\mathrm{H}}_2\mathrm{O}& \\ & & \end{array}$

$\downarrow$ $\downarrow$

6ogm 180gm

240m solution

240 gm solution $=200\mathrm{ml}$ solutution