Massa relatif molekul ($M_{r}$) merupakan jumlah massa relatif atom ($A_{r}$) dari unsur penyusun molekul.

$\begin{aligned}M_{r} & =\Sigma A_{r}\\ M_{r}\mbox{ H}_{2}\mbox{SO}_{4} & =(2\times A_{r}\mbox{ H})+(1\times A_{r}\mbox{ S})+(4\times A_{r}\mbox{ O})\\ & =(2\times1)+(1\times32)+(4\times16)\\ & =2+32+64\\ & =98. \end{aligned}$

Hubungan mol dengan massa ditunjukkan dengan

$n=\frac{m}{A_{r}\mbox{atau}M_{r}}$

$n$ = mol

$m$ = massa

$A_{r}$= massa relatif atom

$M_{r}$= massa relatif molekul

Dari soal diketahui:

massa $\mbox{CO}_{2}$= 11 gram

$\begin{aligned}M_{r}\mbox{CO}_{2} & =(1\times A_{r}\mbox{ C})+(2\times A_{r}\mbox{ O})\\ & =(1\times12)+(2\times16)\\ & =12+32\\ & =44\\ \mbox{mol}\mbox{CO}{}_{2} & =\frac{11}{44}\\ & =0,25\,\mbox{mol.} \end{aligned}$

$\begin{aligned}\mbox{mol} & =\frac{\mbox{jumlah molekul}}{6,02\times10^{23}}\\ & =\frac{1,204\times10^{24}}{6,02\times10^{23}}\mbox{ mol}\\ & =2\mbox{ mol}. \end{aligned}$

(Vm = volume 1 mol gas pada kondisi RTP = 24 L)

$\begin{aligned}\mbox{volume} & =\mbox{mol}\times\mbox{Vm}\\ & =0,02\times24\mbox{ L}\\ & =0,48\mbox{ L}. \end{aligned}$

$\begin{aligned}M & =\frac{n}{V}\\ & =\frac{m}{M_{r}}\times\frac{1000}{V}\\ & =\frac{0,4}{40}\times\frac{1000}{250}\\ & =0,04\mbox{ M}. \end{aligned}$

$\begin{aligned}M_{r}\mbox{ X}_{2}\mbox{O}_{5} & =(2\times A_{r}\mbox{ X})+(5\times A_{r}\mbox{ O})\\ 108 & =2A_{r}\mbox{ X}+(5\times16)\\ 108 & =2A_{r}\mbox{ X}+80\\ 2A_{r}\mbox{ X} & =108-80\\ 2A_{r}\mbox{ X} & =28\\ A_{r}\mbox{ X} & =\frac{28}{2}\\ A_{r}\mbox{ X} & =14\\ M_{r}\mbox{ XCl}_{3} & =(1\times A_{r}\mbox{ X})+(3\times A_{r}\mbox{ Cl})\\ & =(1\times14)+(3\times35,5)\\ & =14+106,5\\ & =120,5. \end{aligned}$

a.) massa dari 0,02 mol NaOH

$\begin{aligned}\mbox{massa NaOH} & =\mbox{mol}\times M_{r}\mbox{ NaOH}\\ & =0,20\times((1\times A_{r}\mbox{ }\mbox{Na})+(1\times A_{r}\mbox{ O})+(1\times A_{r}\mbox{ H})\\ & =0,20\times40\mbox{ gram}\\ & =8\mbox{ gram} \end{aligned}$

b.) massa dari 0,125 mol $\mbox{CH}_{3}\mbox{COOH}$

$\begin{aligned}\mbox{massa}\mbox{CH}_{3}\mbox{COOH} & =\mbox{mol}\times M_{r}\mbox{CH}_{3}\mbox{COOH}\\ & =0,125\times(2\times A_{r}\mbox{ C})+((4\times A_{r}\mbox{ H})+(2\times A_{r}\mbox{ O}))\\ & =0,125\times60\mbox{ gram}\\ & =7,5\mbox{ gram} \end{aligned}$

c.) massa dari 0,05 mol KI

$\begin{aligned}\mbox{massa} & =\mbox{mol}\times M_{r}\mbox{KI}\\ & =0,05\times(39+127)\mbox{ gram}\\ & =8,3\mbox{ gram} \end{aligned}$

d.) massa dari 0,13 mol $\mbox{HNO}_{3}$

$\begin{aligned}\mbox{massa} & =\mbox{mol}\times M_{r}\mbox{HNO}_{3}\\ & =0,13\times((1\times A_{r}\mbox{ H})+(1\times A_{r}\mbox{ N})+(3\times A_{r}\mbox{O}))\\ & =0,13+(1+14+48)\mbox{ gram}\\ & =8,19\mbox{ gram}. \end{aligned}$

e.) massa dari 0,15 mol KOH

$\begin{aligned}\mbox{massa} & =\mbox{mol}\times M_{r}\mbox{KOH}\\ & =0,15\times((1\times A_{r}\mbox{ K})+(1\times A_{r}\mbox{ O})+((1\times A_{r}\mbox{H}))\\ & =0,15\times56\mbox{ gram}\\ & =8,4\mbox{ gram}. \end{aligned}$

$\mbox{CaCl}_{2}\rightarrow\mbox{Ca}^{2+}\mbox{+2Cl}^{-}$

Jumlah ion dalam satu molekul $1\mbox{Ca}^{2+}+\mbox{2Cl}^{-}=3\mbox{ }\mbox{ion}$

$\begin{aligned}\mbox{Jumlah ion dalam 0,02 mol} & =\mbox{jumlah molekul}\times\mbox{jumlah ion dalam satu molekul}\\ & =(0,02\times6,02\times10^{23})\times3\mbox{ ion}\\ & =3,612\times10^{22}\mbox{ion.} \end{aligned}$

$\begin{aligned}V & =n\times22,4\mbox{ L}\\ & =\frac{5,35}{53,5}\times22,4\mbox{L}\\ & =0,1\times22,4\mbox{L}\\ & =2,24\mbox{ L.} \end{aligned}$

$\begin{aligned}M & =\frac{n}{V}\\ & =\frac{\frac{6}{24}}{0,5}\mbox{ M}\\ & =0,5\mbox{ M}. \end{aligned}$

$\begin{alignedat}{1}\mbox{n XO}_{2} & =\frac{5,6\mbox{ L}}{22,4\mbox{ L}}\mbox{ mol}\\ & =0,25\mbox{ mol}\\ M_{r}\mbox{ XO}_{2} & =\frac{11,5}{0,25}\\ & =46\\ M_{r}\mbox{ XO}_{2} & =(1\times A_{r}\mbox{ X})+(2\times A_{r}\mbox{ O})\\ 46 & =A_{r}\mbox{ X}+(2\times16)\\ A_{r}\mbox{ X} & =46-32\\ & =14 \end{alignedat}$

Unsur dengan nilai $A_{r}$ = 14 adalah nitrogen (N).

$\begin{alignedat}{1}\mbox{m larutan NaCl} & =1,1\times200\mbox{ gram}\\ & =220\mbox{ gram}\\ \mbox{m NaCl dalam larutan} & =10\%\times220\mbox{ gram}\\ & =22\mbox{ gram}\\ \mbox{n NaCl} & =\frac{22}{58,5}\mbox{ mol}\\ & =0,376\mbox{ mol}\\ \Sigma\mbox{molekul NaCl} & =0,376\times6,02\times10^{23}\mbox{molekul}\\ & =2,26\times10^{23}\mbox{molekul}\\ \mbox{NaCl}(aq) & \rightarrow\mbox{Na}^{+}(aq)+\mbox{Cl}^{-}(aq)\\ \Sigma\mbox{ion} & =2\times2,26\times10^{23}\mbox{ion}\\ & =4,52\times10^{23}\mbox{ion}. \end{alignedat}$

$\begin{alignedat}{1}n_{\mbox{AB}_{2}} & =\frac{1,15}{A+(16\times2)}\\ & =\frac{1,15}{A+32}\\ V_{\mbox{AB}_{2}} & =n_{\mbox{AB}_{2}}\times V_{M}\\ 0,6\mbox{ dm}^{3} & =\frac{1,15}{A+32}\times24\mbox{ dm}^{3}\\ \mbox{A} & =14\\ V_{\mbox{A}_{2}\mbox{B}_{5}} & =n_{\mbox{A}_{2}\mbox{B}_{5}}\times V_{M}\\ & =\frac{2,16}{(2\times14)+(5\times16)}\times24\mbox{ dm}^{3}\\ & =0,48\mbox{ dm}^{3}. \end{alignedat}$

$\begin{alignedat}{1}n_{\mbox{CaCl}_{2}} & =\frac{1,665}{111}\mbox{ mol}\\ & =0,015\mbox{ mol}\\ M_{\mbox{awal}} & =\frac{n}{V}\\ & =\frac{0,015}{0,25}\mbox{ M}\\ & =0,06\mbox{ M} \end{alignedat}$

$\begin{array}{c} \mbox{CaCl}_{2}\\ \frac{1}{2}\times0,03\mbox{M}\\ 0,015\mbox{M} \end{array}$$\begin{array}{ccc} \rightarrow & \mbox{Ca}^{2+} & +\mbox{2Cl}^{-}\\ & & 0,03\mbox{ M}\\ \\ \end{array}$

$\begin{alignedat}{1}M_{1}\cdot V_{1} & =M_{2}\cdot V_{2}\\ 0,06\times0,25 & =0,015\times V_{2}\\ V_{2} & =1\mbox{ L}\\ V_{\mbox{ditambah}} & =1\mbox{ L}-0,25\mbox{ L}\\ & =0,75\mbox{ L}\\ & =750\mbox{ mL}. \end{alignedat}$

$\begin{alignedat}{1}V_{\mbox{RTP}} & =V_{M}\times n\\ 0,6\mbox{ L} & =24\mbox{ L}\times\frac{1,775}{M_{r}\mbox{ X}_{2}}\\ M_{r}\mbox{ X}_{2} & =71\\ A_{r}\mbox{ X} & =35,5. \end{alignedat}$