[અણુ દળ - ${Na}: 23.0\, {u}, {O}: 16.0\, {u}, {P}: 31.0 \,{u}]$
$\frac{1}{3} \times \frac{3.45}{23} {~mol} \quad$ $3.45 {~g}$
$\quad\quad\quad\quad\quad\quad\quad\frac{3.45}{23} {~mol}$
therefore molarity of ${Na}_{3} {PO}_{4}$ Solution =
$\frac{{n}_{{Na}_{3} {PO}_{4}}}{\text { volume of solution in } {L}}$
$=\frac{\frac{1}{3} \times \frac{3.45}{23}\, {~mol}}{0.1\, {~L}}$
$=0.5=50\, \times 10^{-2}$
પદાર્થ |
$\% Sn$ |
$\% O$ |
A |
78.77 |
21.23 |
B |
88.12 |
11.88 |