1. Tardigrade
  2. Question
  3. Chemistry
  4. H 3 PO 4( d =1.8 g / mL ) is 18 M. Hence, mass percentage and molality are:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $H _{3} PO _{4}( d =1.8\, g\, / mL )$ is $18 \,M$. Hence, mass percentage and molality are:

Some Basic Concepts of Chemistry

Solution:

Density $=1.8 \,g /\, mL$

Molarity $=18 \,M =18\, mol \,H _{3} PO _{4}$ in $1\, L$ solution

$1000 \,mL$ solution has $H _{3} PO _{4}=18 \,mol =18 \times 98 \,g$

$100 \,mL =(1800 \,g )$ solution has $H _{3} PO _{4}=18 \times 98\, g$

$H _{3} PO _{4}$ by mass $\%$ (by weight of solution)

$=\frac{18 \times 98}{1800} \times 100=98 \%$

or $(1800-18 \times 98) \,g\, H _{2} O$ (solvent) $H _{3} PO _{4}=18 \times 98\, g$

$2\, g$ solvent has $H _{3} PO _{4}=98\, g =1 \,mol \,H _{3} PO _{4}$

$1 \,g =1000 \,g$ solvent has $H _{3} PO _{4}=500$ molal

Thus, $98 \%, 500$ molal