### Related Formula M = frac\% text purity times textdensity times 10textMolar Mass$M = \frac{\% \text{ purity} \times \text{density} \times 10}{\text{Molar Mass}}$ ### Core Logic Specific gravity is numerically equivalent to density in textg/cm^3$\text{g/cm}^3$, so density = 1.54text g/mL$1.54\text{ g/mL}$. Volume of solution = 1text L = 1000text mL$1\text{ L} = 1000\text{ mL}$. Mass of solution = textVolume times textDensity = 1000 times 1.54 = 1540text g$\text{Volume} \times \text{Density} = 1000 \times 1.54 = 1540\text{ g}$. ### Step 1: Finding Solute Mass and Molarity Since the purity is 70\%$70\%$ by weight, the mass of H_3PO_4$H_3PO_4$ in the solution is: textMass of H_3PO_4 = 1540 times 0.70 = 1078text g$\text{Mass of } H_3PO_4 = 1540 \times 0.70 = 1078\text{ g}$. Moles of H_3PO_4 = frac107898 = 11text moles$H_3PO_4 = \frac{1078}{98} = 11\text{ moles}$. Since this is dissolved in 1text L$1\text{ L}$ of solution, the Molarity is: M = frac11text moles1text L = 11text M$M = \frac{11\text{ moles}}{1\text{ L}} = 11\text{ M}$ ### Pattern Recognition Shortcut formula directly substitutes the values: M = frac70 times 1.54 times 1098 = frac107898 = 11$M = \frac{70 \times 1.54 \times 10}{98} = \frac{1078}{98} = 11$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions Class 11 Chemistry: Some Basic Concepts of Chemistry