Question

Calculate normality, molarity and molality of the solution containing $22 \%( w / w )$ salt $Al _{2}\left( SO _{4}\right)_{3}\{ d =1.253 gm / ml \}$ by weight is

• A. $4.83 \, N, \, 8.25 \, M, \, 8.25 \, m$
• B. $48.3 \, N, \, 0.825 \, M, \, 0.825 \, m$
• C. $4.83 \, N, \, 4.83 \, M, \, 4.83 \, m$
• D. $4.83 \, N, \, 0.805 \, M,0.825 \, m$

Answer 1

Answer: (D)

Normality $=\frac{w / w \, \% \times \text{10} \times d}{E_{2}}$ $=\frac{\text{22} \times \text{10} \times \text{1.253}}{\frac{\text{342}}{6}}=\text{4.83 }N$

Molarity $=\frac{w / w \, \% \times \text{10} \times d}{M_{2}}=\frac{\text{22} \times \text{10} \times \text{1.253}}{\text{342}}=\text{0.805 }M$

$E_{2}= \, $ eq.mass of solute

$M_{2}= \, $ molar ass of solute

$Molality \, =\frac{\left(w / w \%\right) \times 1000}{M_{2} \left(100 - w / w \%\right)}$

Molality $=\frac{\text{22} \times \text{1000}}{\text{342} \left(\right. \text{100} - \text{22} \left.\right)}=\text{0.825 }m$