Question

1.Calculate the molarity of a solution containing 2.40 g of sodium chloride (NaCl) and 40mL solution.
2.What is the molality of 60.5% by mass of nitric acid (HNO3) solution?
3.Calculate the mole fraction of HCl and H2O in a 20% (w/w) aqueous HCl solution.

Answer 1

1. The molarity will be

$$C=\frac{n_\text{NaCl}}{V}=\frac{m/\Mu}{V}=\frac{2.4/(23+35.5)}{40}=1.03\text{ mol/mL}.$$

2. The molarity, on the other hand, can be defined in terms of mass:

$$m_{water}=m_{total}-m_{acid}.$$

$$C_m=\frac{m_{acid}/\Mu}{m_{water}}=\frac{m_{acid}/\Mu}{m_{acid}(1/0.605-1)}=\frac{1/0.605-1}{\Mu}=\\\space\\ =0.01\text{ mol/g}.$$

3. Mass of water can be defined in terms of solute mass m:

$$m_w=m(1-0.2)=0.8m$$

Find the amount of substance:

$$n_w=m_w/\Mu_w=0.8m/\Mu_w,\\ n_h=m_h/\Mu_h=0.2m/\Mu_c.$$

Find the mole fraction:

$$\chi_w=\frac{n_w}{n_w+n_h}=\frac{0.8m/\Mu_w}{0.8m/\Mu_w+0.2m/\Mu_h}=\\\space\\ =\frac{0.8/18}{0.8/18+0.2/36.5}=0.89.\\\space\\ \chi_h=\frac{n_h}{n_w+n_h}=\frac{0.2m/\Mu_h}{0.8m/\Mu_w+0.2m/\Mu_h}=\\\space\\ =\frac{0.2/36.5}{0.8/18+0.2/36.5}=0.11.$$