0.2 Molar concentration  (Page 3/3)

Problem : In a container, 14 gm nitrogen is mixed with 2 gm of hydrogen gas to form ammonia gas. The amount of ammonia formed is 5.1 gm. Determine mole fractions of ammonia in the container.

Solution : Here, moles of nitrogen and hydrogen present in the container are :

$$\text{Moles of nitrogen}=\frac{14}{2X14}=0.5$$

$$\text{Moles of hydrogen}=\frac{2}{1X2}=1$$

In order to determine mole fraction, we need to know the moles of gases remaining after the reaction. We can know the moles of gas provided we know the extent reaction takes place. The mass of product is known here. Using chemical equation, we can determine the moles of nitrogen and hydrogen used.

$$\text{Moles of ammonia produced}=\frac{5.1}{\left(14+3X1\right)}=\frac{5.1}{17}=0.3$$

The chemical equation is :

$$N_2+3H_2\to 2N{H}_3$$

Applying mole concept :

$$\text{1 mole of}{N}_2\equiv \text{3 moles of}{H}_2\equiv \text{2 moles of}N{H}_3$$

Thus, moles of nitrogen and hydrogen gas consumed are :

$$\text{Moles of}{N}_2\text{consumed}=\frac{1}{2}X0.3=0.15$$

$$\text{Moles of}{H}_2\text{consumed}=\frac{3}{2}X0.3=0.45$$

Clearly, reaction does not exhaust either of gases. The moles remaining in the container after reaction are :

$$\text{Moles of}{N}_2\text{remaining}=0.5-0.15=0.35$$

$$\text{Moles of}{H}_2\text{remaining}=1.0-0.45=0.55$$

$$\text{Moles of}N{H}_3\text{produced}=0.3$$

$$\text{Mole fraction of}N{H}_3=\frac{0.3}{0.3+0.35+0.55}=\frac{0.3}{1.2}=0.25$$

Problem : Determine (i) molarity of 750 cc NaOH solution containing 40 gm NaOH (ii) moles and milli-moles of NaOH in 500 ml of 0.4 M NaOH solution and (iii) mass of NaOH in 500 ml of 0.2 M NaOH solution.

Solution : (i) In the first part, volume is given in cc. Hence, we use the formula :

$$\Rightarrow M=\frac{n_B}{V_{CC}}X1000$$

In order to use this formula, we need to find the moles present. Here, molecular weight of NaOH is 23+16+1 = 40. Hence,

$$n_B=40/40=1$$

Putting values in the expression, we have :

$$\Rightarrow M=\frac{1}{750}X1000=1.33M$$

(ii) Moles of solute is :

$$\Rightarrow n_B=\frac{0.4X500}{1000}=0.2$$

$$\Rightarrow \text{Millimoles of solute}=M{V}_{CC}=0.4X500=200$$

(iii) Moles of solute is :

$$\Rightarrow n_B=\frac{0.2X500}{1000}=0.1$$

The mass of NaOH is :

$$\Rightarrow g_B=n_{B}X{M}_{NaOH}=0.1X40=4gm$$

Problem : Two litres of a solution is prepared, which contains 44.4 gm of calcium hydro-oxide. Find milli-moles of calcium hydro-oxide in 100 ml sample of the solution.

Solution : Here we first need to determine the molarity of the solution as :

$$M=\frac{n_B}{V_L}$$

The moles of calcium hydro-oxide is :

$$\Rightarrow n_B=\frac{44.4}{M_{Ca{\left(OH\right)}_2}}=\frac{44.4}{40+2X\left(16+1\right)}=\frac{44.4}{74}=0.6$$

Putting this in the formula, we have :

$$\Rightarrow M=\frac{n_B}{V_L}=\frac{0.6}{2}=0.3M$$

The 100 ml sample has the same molarity as that of the bulk solution. Hence,

$$\Rightarrow \text{Milli-moles}=M{V}_{CC}=0.3X100=30$$

Problem : What is molarity of pure water assuming its density 1 gm/cc ?

Solution : Pure water is one component system. According to definition, molarity is given as :

$$M=\frac{n_B}{V_L}$$

Let $V_L$ = 1 litre. The mass of 1 litre water is 1000 gm as density is 1 gm/cc. The number of moles in 1000 gm is :

$$n_B=\frac{1000}{18}=55.56$$

Putting this value in the expression of molarity, we have :

$$\Rightarrow M=\frac{n_B}{V_L}=\frac{55.56}{1}=55.56$$

Problem : Sulphuric acid solution of density 1.8 gm/cc contains 24.5% acid by weight. What is molarity of the solution?

Solution : Here,

$$\text{Mass percentage},x=24.5$$

$$\text{Density in gm/cc},d=1.8$$

$$\text{Molecular weight of}{H}_{2}S{O}_4=2X1+32+4X16=98$$

Molarity of the solution is :

$$\Rightarrow M=\frac{10xd}{M_O}=\frac{10X24.5X1.8}{98}=4.5M$$