Question

36.5 (%w/w) $HCl$ solution has density equal to $1.2gm{L}^{-1}$ then :

• A. Its molarity is $12M$
• B. Its molality is $15.7m$
• C. The mole fraction of solute is $0.21$
• D. Its molality is $0.36m$

Answer 1

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

The correct options are
A Its molarity is $12M$
B Its molality is $15.7m$
C The mole fraction of solute is $0.21$

$M=\frac{\%\text{by weight}\times 10\times d}{M.w}=\frac{36.5\times 10\times 1.2}{36.5}=12Mm=\frac{\%\text{by weight}\times 1000}{\text{Molar mass of HCl}\times (100-\%\text{by weight})}=\frac{36.5\times 1000}{36.5(100-36.5)}=\frac{1000}{63.5}=15.7m$

Again, let mass of solution be $100g$
So the mass of $HCl$ would be $36.5g$
$\therefore$ Moles of $HCl=\frac{36.5}{36.5}=1mol$
$\text{moles of water}=\frac{100-36.5}{18}=\frac{64.5}{18}=3.58\approx 3.6$
$\text{mole fraction}=\frac{moles}{\text{total number of moles}}=\frac{1}{4.6}=0.21$