The density of 3 molal solution of NaOH is $1.110 \mathrm{~g} \mathrm{~mL}^{-1}$. Calculate the molarity of the solution.
3 molal solution of NaOH means 3 moles of NaOH are dissolved in 1 kg solvent. So, the mass of solution $=1000 \mathrm{~g}$ solvent $+120 \mathrm{~g} \mathrm{NaOH}=1120 \mathrm{~g}$ solution
(Molar mass of $\mathrm{NaOH}=23+16+1=40 \mathrm{~g}$ and 3 moles of $\mathrm{NaOH}=3 \times 40=120 \mathrm{~g}$ )
$$\text { Volume of solution }=\frac{\text { Mass of solution }}{\text { Density of solution }}\quad \left(\mathrm{Q} d=\frac{m}{V}\right)$$
$$V=\frac{1120 \mathrm{~g}}{1.110 \mathrm{~g} \mathrm{~mL}^{-1}}=1009 \mathrm{~mL}$$
$$\begin{aligned} \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{3 \times 1000}{1009}=2.973 \mathrm{M} \approx 3 \mathrm{M} \end{aligned}$$
Volume of a solution changes with change in temperature, then what will the molality of the solution be affected by temperature? Give reason for your answer.
No, molality of solution does not change with temperature since mass remains unaffected with temperature.
$$\text { Molality, } m=\frac{\text { moles of solute }}{\text { weight of solvent (in } \mathrm{g})} \times 1000$$
If 4 g of NaOH dissolves in 36 g of $\mathrm{H}_2 \mathrm{O}$, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is $1 \mathrm{~g} \mathrm{~mL}^{-1}$ ).
Number of moles of NaOH,
$$\begin{gathered} n_{\mathrm{NaOH}}=\frac{4}{40}=0.1 \mathrm{~mol} \quad \left\{\mathrm{Q} n=\frac{\text { Mass }(\mathrm{g})}{\operatorname{Molar} \text { mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}\right\}\\ \text{Similarly,}\quad n_{\mathrm{H}_2 \mathrm{O}}=\frac{36}{18}=2 \mathrm{~mol} \end{gathered}$$
$$\begin{gathered} \text { Mole fraction of } \mathrm{NaOH}, X_{\mathrm{NaOH}}=\frac{\text { moles of } \mathrm{NaOH}}{\text { moles of } \mathrm{NaOH}+\text { moles of } \mathrm{H}_2 \mathrm{O}} \\ \qquad X_{\mathrm{NaOH}}=\frac{0.1}{0.1+2}=0.0476 \end{gathered}$$
$$\begin{aligned} &\text { Similarly, }\\ &\begin{aligned} X_{\mathrm{H}_2 \mathrm{O}} & =\frac{n_{\mathrm{H}_2 \mathrm{O}}}{n_{\mathrm{NaOH}}+n_{\mathrm{H}_2 \mathrm{O}}} \\ & =\frac{2}{0.1+2}=0.9524 \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { Total mass of solution } & =\text { mass of solute }+ \text { mass of solvent } \\ & =4+36=40 \mathrm{~g} \\ \text { Volume of solution } & =\frac{\text { Mass of solution }}{\text { specific gravity }}=\frac{40 \mathrm{~g}}{1 \mathrm{~g} \mathrm{~mL}^{-1}}=40 \mathrm{~mL} \\ \text { Molarity } & =\frac{\text { Moles of solute } \times 1000}{\text { Volume of solution }(\mathrm{mL})} \\ & =\frac{0.1 \times 1000}{40}=2.5 \mathrm{M} \end{aligned}$$
The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2 A+4 B \rightarrow 3 C+4 D$, when 5 moles of $A$ react with 6 moles of $B$, then
(a) which is the limiting reagent?
(b) calculate the amount of $C$ formed?
$$2 A+4 B \longrightarrow 3 C+4 D$$
According to the given reaction, 2 moles of $A$ react with 4 moles of $B$.
Hence, 5 moles of $A$ will react with 10 moles of $B\left(\frac{5 \times 4}{2}=10\right.$ moles $)$
(a) It indicates that reactant $B$ is limiting reagent as it will consume first in the reaction because we have only 6 moles of $B$.
(b) Limiting reagent decide the amount of product produced.
According to the reaction,
4 moles of $B$ produces 3 moles of $C$
$\therefore 6$ moles of $B$ will produce $\frac{3 \times 6}{4}=4.5$ moles of $C$.
Match the following.
| A. | 88 g of CO$$_2$$ | 1. | 0.2 mol |
|---|---|---|---|
| B. | $$6.022\times10^{23}$$ molecules of H$$_2$$O | 2. | 2 mol |
| C. | 5.6 L of O$$_2$$ at STP | 3. | 1 mol |
| D. | 96 g of O$$_2$$ | 4. | $$6.022\times10^{23}$$ molecules |
| E. | 1 mole of any gas | 5. | 3 mol |
A. $\rightarrow(2)$
B. $\rightarrow(3)$
C. $\rightarrow$ (1)
D. $\rightarrow$ (5)
E. $\rightarrow(4)$
$$\text { A. Number of moles of } \mathrm{CO}_2 \text { molecule }=\frac{\text { Weight in gram of } \mathrm{CO}_2}{\text { Molecular weight of } \mathrm{CO}_2}=\frac{88}{44}=2 \mathrm{~mol}$$
$$\begin{aligned} \text { B. } \quad 1 \text { mole of a substance } & =N_A \text { molecules }=6.022 \times 10^{23} \text { molecules } \\ & =\text { Avogadro number } \\ & =6.022 \times 10^{23} \text { molecules of } \mathrm{H}_2 \mathrm{O}=1 \mathrm{~mol} \end{aligned}$$
$$\begin{aligned} \text { C. } 22.4 \mathrm{~L} \text { of } \mathrm{O}_2 \text { at } \mathrm{STP} & =1 \mathrm{~mol} \\ 5.6 \mathrm{~L} \text { of } \mathrm{O}_2 \text { at } \mathrm{STP} & =\frac{5.6}{22.4} \mathrm{~mol}=0.25 \mathrm{~mol} \end{aligned}$$
D. Number of moles of 96 g of $\mathrm{O}_2=\frac{96}{32} \mathrm{~mol}=3 \mathrm{~mol}$
E. 1 mole of any gas $=$ Avogadro number $=6.022 \times 10^{23}$ molecules