Question

$H_{3}P{O}_4(d=1.8g/mL)$ is $18M$. Hence, mass percentage and molality are:

• A. 18,32.4
• B. 98,32.4
• C. 98,500
• D. 98,18

Answer 1

Answer: (C)

Density $=1.8g/mL$

Molarity $=18M=18mol{H}_{3}P{O}_4$ in $1L$ solution

$1000mL$ solution has $H_{3}P{O}_4=18mol=18\times 98g$

$100mL=(1800g)$ solution has $H_{3}P{O}_4=18\times 98g$

$H_{3}P{O}_4$ by mass $\%$ (by weight of solution)

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

or $(1800-18\times 98)g{H}_{2}O$ (solvent) $H_{3}P{O}_4=18\times 98g$

$2g$ solvent has $H_{3}P{O}_4=98g=1mol{H}_{3}P{O}_4$

$1g=1000g$ solvent has $H_{3}P{O}_4=500$ molal

Thus, $98\%,500$ molal