Chemistry MCQs for NEET — Practice Questions with Answers

100 mL gaseous mixture contain 20 mL $C_3{H}_8$

So, volume of $C{H}_4$ and CO = (100 - 20) - 80 mL

$$\begin{gathered} C_3{H}_8 + 5O_2 \to 3C{O}_2 + 4H_{2}O \\ C{H}_4 + 2O_2 \to C{O}_2 + 2H_{2}O; \\ CO + \frac{1}{2}{O}_2 \to C{O}_2 \end{gathered}$$

80 mL ($C{H}_4$ and CO) will produce 80 mL $C{O}_2;C_3{H}_8$ will produce = 3 X 20 = 60 mL

Total $C{O}_2$ produce= 80+60 = 140

$\% \mathrm{of} \mathrm{O}=\frac{16\times 27}{\left(100+3\times 310\right)}\times 100 = 41.94\%$

Moles of Ag₃A= $\frac{moles of Ag}{3}$

=$$\begin{gathered} \frac{0.607}{M}= \frac{0.37}{108}\times \frac{1}{3}\Rightarrow M= 531 \\ \therefore mol. wt. of H_{3}A =mol. wt. of A{g}_{3}A - 3\times At. wt. of Ag + 3\times At. wt. of H= 210 \end{gathered}$$

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{PM}}{\mathrm{RT}} \Rightarrow \mathrm{M}=\frac{\mathrm{dRT}}{\mathrm{P}}=\frac{2.28\times 0.0821\times 300}{1}=56.15 \mathrm{g}/\mathrm{mol} \\ \mathrm{E}.\mathrm{F}.=\frac{85.7}{12}:\frac{14.3}{1}=7.14 : 14.3=1 : 2; \\ \therefore \mathrm{E}.\mathrm{F}. \mathrm{is} {\mathrm{CH}}_2; \mathrm{M}.\mathrm{F}.={\left({\mathrm{CH}}_2\right)}_{\mathrm{n}} \\ \mathrm{where} \mathrm{n}=\frac{56.15}{12+2}\simeq 4 \\ \therefore \mathrm{M}.\mathrm{F}. \mathrm{is} {\mathrm{C}}_4{\mathrm{H}}_8 \end{gathered}$$

Percentage above 100 represents the mass of ${\mathrm{H}}_2\mathrm{O}$ that reacts with dissolved ${\mathrm{SO}}_3$ in oleum to give ${\mathrm{H}}_2{\mathrm{SO}}_4$, i.e., 9 g ${\mathrm{H}}_2\mathrm{O}$ reacts with free ${\mathrm{SO}}_3$ to produce ${\mathrm{H}}_2{\mathrm{SO}}_4$

     ${\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_3 \to {\mathrm{H}}_2{\mathrm{SO}}_4$

     18 g ${\mathrm{H}}_2\mathrm{O}$ reacts with 80 g ${\mathrm{SO}}_3$

$\therefore$ 9 g ${\mathrm{H}}_2\mathrm{O}$ will react with 40 g ${\mathrm{SO}}_3$

or  % of free ${\mathrm{SO}}_3$ = 40

$$\begin{gathered} \mathrm{Mol}. \mathrm{wt}. \mathrm{of} {\mathrm{XY}}_2=\frac{5}{0.05}=100 \\ \mathrm{Mol}. \mathrm{wt}. \mathrm{of} {\mathrm{X}}_2{\mathrm{Y}}_3=\frac{85}{3.011\times {10}^{23}}\times {\mathrm{N}}_{\mathrm{A}}=170 \\ \mathrm{Let} \mathrm{molar} \mathrm{mass} \mathrm{of} \mathrm{X} \mathrm{and} \mathrm{Y} \mathrm{are} \mathrm{a} \mathrm{and} \mathrm{b} \mathrm{respectively} \\ \therefore \mathrm{a}+2\mathrm{b}=100 \\ 2\mathrm{a}+3\mathrm{b}=170; \\ \mathrm{a}=40; \\ \mathrm{b}=30 \end{gathered}$$

$$\begin{gathered} \frac{4.92}{1000}\times 1 = \frac{40\times {10}^{-3}}{\mathrm{M}}\times 0.0821\times 300; \\ \mathrm{M}\simeq 200; \mathrm{Atomic} \mathrm{mass} \mathrm{of} \mathrm{X}=100 \end{gathered}$$

$$\begin{gathered} 2\mathrm{Fe}{\left({\mathrm{NO}}_3\right)}_3 + 3{\mathrm{Na}}_2{\mathrm{CO}}_3 \stackrel{}{\to } {\mathrm{Fe}}_2{\left({\mathrm{CO}}_3\right)}_3+6{\mathrm{NaNO}}_3 \\ \mathrm{mole} 2.5 3.6 \\ \mathrm{mole}/\mathrm{stoichiometric} \mathrm{coefficient} 1.25 1.2 \\ \mathrm{Limiting} \mathrm{reagent} \mathrm{is} {\mathrm{Na}}_2{\mathrm{CO}}_3 \mathrm{so} \mathrm{moles} \mathrm{of} {\mathrm{NaNO}}_3 \mathrm{should} \mathrm{be} \mathrm{formed} = 3.6\times 2=7.2 \\ \% \mathrm{yield} = \frac{6.3}{7.3}\times 100=87.5 \end{gathered}$$

$$\begin{gathered} {\mathrm{CO}}_2+2\mathrm{NaOH} \stackrel{}{\to } {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{n}}_{\mathrm{NaOH}}=1; \therefore {\mathrm{CO}}_2 \mathrm{present} \mathrm{in} \mathrm{mixture}=0.5 \mathrm{and} \mathrm{Co} \mathrm{present}=0.3 \mathrm{mole} \\ \mathrm{When} \mathrm{more} {\mathrm{CO}}_2 \mathrm{produced}=0.3, \mathrm{more} \mathrm{NaOH} \mathrm{requried}=0.3\times 2=0.6 \mathrm{mole} \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{reaction} \mathrm{that} \mathrm{takes} \mathrm{place} \mathrm{is} \\ \mathrm{NaCl}+{\mathrm{AgNO}}_3 \stackrel{}{\to } \mathrm{AgCl}\downarrow +{\mathrm{NaNO}}_3 \\ \because 143.5 \mathrm{g} \mathrm{of} \mathrm{AgCl} \mathrm{is} \mathrm{produced} \mathrm{from} 58.5 \mathrm{g} \mathrm{NaCl} \\ \therefore 14 \mathrm{g} \mathrm{of} \mathrm{AgCl} \mathrm{will} \mathrm{produce} \mathrm{from} \\ \frac{58.5\times 14}{143.5}=5.70\mathrm{g} \\ \mathrm{This} \mathrm{is} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{NaCl} \mathrm{in} \mathrm{common} \mathrm{slat}; \\ \% \mathrm{purity}=\frac{5.70}{6}\times 100=95\% \end{gathered}$$