Some Basic Concepts in Chemistry MCQs for NEET — Chemistry Questions with Answers

Balanced reaction is

$2F_2{O}_3 + 3C \to 4Fe + 3C{O}_2$

No. of moles of $F{e}_2{O}_3=\left(\frac{120\times 1000}{2\times 56+48}\right)\times \left(\frac{90}{100}\right)$

Mass of 80% pure iron produced

$=\frac{120\times 1000\times 0.9}{2\times 56+48}\times \frac{2\times 56}{0.8}$

= 94500 gram or 94.5 kg

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{CO}}_2 \mathrm{in} 100 \mathrm{g} \mathrm{mixture} =\frac{66}{44} = 1.5 \\ \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{H}}_2 \mathrm{in} 100 \mathrm{g} \mathrm{mixture} = \frac{34}{2} = 17 \\ {\mathrm{M}}_{\mathrm{average}} = \frac{1000}{18.5}=5.40 \\ \mathrm{V}.\mathrm{D}. = \frac{5.4}{2} = 2.7 \end{gathered}$$

Let 1 mole of mixture has x mole ${\mathrm{N}}_2{\mathrm{O}}_4$

$2\times 27.6=\mathrm{x} \left(92\right)+\left(1-\mathrm{x}\right) 46; \mathrm{x}=0.2$

Let V mL of alcohol be required

$\because$ mass of alcohol is same in both solutions

$\therefore$ $$\begin{gathered} \frac{75}{100}\times 0.8\times V= \frac{30}{100}\times 0.9\times 150 \\ V= 67.5 mL \end{gathered}$$

$$\begin{gathered} \mathrm{Let} \mathrm{mole} \% \mathrm{of} {}^{26}\mathrm{Mg} \mathrm{be} \mathrm{x} \\ \therefore \frac{\left(21-\mathrm{x}\right) 25+\mathrm{x}\left(26\right)+79\left(24\right)}{100}=24.31 \\ \mathrm{x}=10\% \end{gathered}$$

$$\begin{gathered} Let wt. of N{H}_{4}N{O}_3 and (N{H}_4{)}_{2}HP{O}_4 are x and y gram respectively \\ \frac{{\displaystyle \frac{x}{80}}\times 2\times 14+{\displaystyle \frac{y}{132}}\times 2\times 14}{x+y}\times 100 = 30.4 \\ \Rightarrow x:y =2 : 1 \end{gathered}$$

$$\begin{gathered} \mathrm{Let} \mathrm{mole} \mathrm{fraction} \mathrm{of} {\mathrm{O}}_2 \mathrm{is} \mathrm{x} \\ 40=32\times \mathrm{x}+80\left(1-\mathrm{x}\right) \\ \mathrm{or} \mathrm{x}=5/6 \\ \mathrm{a} : \mathrm{b}=\mathrm{x} :\left(1-\mathrm{x}\right) = \frac{5}{6} : \frac{1}{6} \\ \mathrm{When} \mathrm{ratio} \mathrm{is} \mathrm{changed} \\ {\mathrm{M}}_{\mathrm{Mixture}} = 32\times \frac{1}{6}+80\times \frac{5}{6}=72 \end{gathered}$$

$$\begin{gathered} \mathrm{Metal} \mathrm{oxide} = 2.74 \mathrm{g}; \\ \mathrm{wt}. \mathrm{of} \mathrm{vanadium} = 1.53 \mathrm{g} \\ \% \mathrm{of} \mathrm{V}=\frac{1.53}{2.74}\times 100=55.83 \\ \mathrm{Thus}, \% \mathrm{of} \mathrm{O}=100-55.83=44.17 \\ \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{V}=\frac{55.83}{52}=1.1 \\ \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{V}=\frac{44.17}{16}=2.76 \\ \mathrm{Simplest} \mathrm{ratio} \mathrm{of} \mathrm{V} \mathrm{and} \mathrm{O}=1:2.5 \mathrm{or} 2:5 \\ \mathrm{Hence}, \mathrm{the} \mathrm{empirical} \mathrm{formula}={\mathrm{V}}_2{\mathrm{O}}_5 \end{gathered}$$

Suppose, the volume of propane = V L

$$\begin{gathered} \therefore C_3{H}_8\left(g\right) + 5O_2\left(g\right) \to 3C{O}_2\left(g\right) + 4H_{2}O\left(l\right) \\ V 5V 3V \\ {C}_4{H}_{10}\left(g\right)+\frac{13}{2}{O}_2\left(g\right) \to 4C{O}_2\left(g\right) + 5H_{2}O\left(l\right) \\ (3-V) \frac{13}{2}(3-V) 4(3-V) \\ \because Total volume of C{O}_{2 }produced = 10 L; \\ \because 3V + 4(3-V) = 11; \therefore V=1 \\ \therefore Volume of bu\tan e = (3-1) = 2L \\ Thus, the ratio of volume of bu\tan e to propane= 2 : 1 \end{gathered}$$

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