Question

$5$ g of NaCl is dissolved in $1000$ g of water. If the density of the resulting solution is $0.997$ g per cc. Which of the following option(s) is/are correct?

• A. Mole fraction of solute is $1.53\times {10}^{-3}$
• B. Molarity of the solution is $0.085M$
• C. Normality of the solution is $0.085N$
• D. Molality of the solution is $0.854m$

Answer 1

Answer: (A), (B), (C)

Normality of the solution is $0.085N$

$\text{Mole of NaCl}=\frac{5}{58.5}=0.0854$ (mol. wt. of NaCl $=58.5$)
By definition:
$\text{Molality}=\frac{\text{moles}}{\text{wt. of solvent in grams}}\times 1000$
$=\frac{0.0854}{1000}\times 1000=0.0854m.$
$\text{Volume of the solution}=\frac{\text{mole}}{\text{density in grams/cc}}=\frac{1005}{0.997}cc$
$=1008mL=1.008L$
Again by definition
$\text{Molarity}=\frac{\text{moles}}{\text{volume of solution in litres}}=\frac{0.0854}{1.008}=0.085M$
$\therefore \text{Normality}=0.085N$
(for NaCl, eq. wt. = mol. wt.)
Further, moles of $H_{2}O=\frac{1000}{18}=55.55\text{(per 1000 g)}$
(Supposing 1 mL = 1 g for water having density 1 g/mL)
Total moles = moles of $NaCl$ + moles of $H_{2}O$
$=0.0854+55.55=55.6409$
$\text{Mole fraction of}NaCl=\frac{\text{moles of NaCl}}{\text{total moles}}=\frac{0.0854}{55.6409}$
$=1.53\times {10}^{-3}$