Number of molecules in 100 mL of each of ${O_2},N{H_3}\;and\;C{O_2}$ at STP are in the order

100 ML ${O_2},N{H_3}\;and\;C{O_2} = \frac{{0.1}}{{22.4}} = \frac{1}{{224}}mol$ For ${O_2}$ no. of molecules $ = \frac{1}{{224}} \times 6.023 \times {10^{23}}$ For $N{H_3}$ no. of molecules $ = \frac{1}{{224}} \times 6.023 \times {10^{23}}$ For $C{O_2}$ no. of molecules$ = \frac{1}{{224}} \times 6.023 \times {10^{23}}$

${\rm{C}}{{\rm{O}}_2} < {{\rm{O}}_2} < N{{\rm{H}}_3}$

${\rm{N}}{{\rm{H}}_3} < {{\rm{O}}_2} < C{{\rm{O}}_2}$

${\rm{N}}{{\rm{H}}_3} = {\rm{C}}{{\rm{O}}_2} < {{\rm{O}}_2}$

The correct Option is "All have same number of molecules"

The number of atoms in 0.1 mol of a triatomic gas is: $\left( {{N_A} = 6.02 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)$

0.1 mole has atoms $ = 0.1 \times 6.02 \times {10^{23}} \times 3$ $ = 1.806 \times {10^{23}}$

The correct Option is "$1.806 \times {10^{23}}$"

Which contains greatest number of oxygen atoms?

16 g O contains N atoms of O 32 g ${O_2}$ contains 2N atoms of O 48 g ${O_3}$ 3 contains 3N atoms of O

The correct Option is "All have the same number of atoms"

Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. Wt. of Fe=56) present in one molecule of haemoglobin is

In 100 g haemoglobin, mass of iron = 0.33 g $\therefore $ in 67200 g haemoglobin, mass of iron $ = \frac{{67200 \times 0.33}}{{100}}$ $\therefore $ the number of Fe atoms in one Hb molecule $ = \frac{{672 \times 0.33}}{{56}}$ $ = 4$

One litre of ${{\rm{C}}{{\rm{O}}_2}}$ is passed over hot coke. The volume becomes 1.4 litre. The per cent composition of products is:

$\mathop {{\rm{C}}{{\rm{O}}_2}}\limits_{\begin{array}{*{20}{c}} 1\\ {\left( {1 - {\rm{x}}} \right)} \end{array}} + \mathop C\limits_0 \to \mathop {2{\rm{CO}}}\limits_{\begin{array}{*{20}{c}} {}\\ {2x} \end{array}} .$ $\therefore 1 - x + 2x = 1.4\;{\rm{find}}\;x$

0.8 litre ${{\rm{C}}{{\rm{O}}_2}}$

The correct Option is "0.6 litre ${{\rm{C}}{{\rm{O}}_2}}$ and 0.8 litre CO"

11.2 litre of ${\rm{N}}{{\rm{H}}_3}\;$ at STP has electrons:

${\rm{1}}\;{\rm{mole}}\;{\rm{N}}{{\rm{H}}_3}\; \equiv {\rm{ }}10{\rm{ }}\;N{\rm{ }}\;electron$ $\frac{{11.2}}{{22.4}}\;mole\;{\rm{N}}{{\rm{H}}_3}\; \equiv {\rm{ }}10{\rm{ }} \times N \times \frac{1}{2} = 3.01 \times {10^{24}}\;electron$

The correct Option is "$3.01 \times {10^{24}}$"

The formula which represents the simple ratio of atoms in a compound is called:

Empirical formula of glucose ${\rm{ = C}}{{\rm{H}}_2}{\rm{O}};$ Molecular formula of glucose $ = {\left( {{\rm{C}}{{\rm{H}}_2}{\rm{O}}} \right)_6}{\rm{.}}$

The correct Option is "Empirical formula"

For the reaction, $X + 2Y \to Z$ 5 Moles of X and 9 moles of Y will produce

In a chemical reaction, coefficient represents mole of that substance. $X + 2Y \to Z$ This indicates 1 mole of X reacts with 2 moles of Y to form 1 mole of Z. So, 5 moles of X will require 10 moles ofY. But we have taken only 9 moles of Y. Hence, Y is in limiting quantity. Hence, we determine product from Y. Thus, 5 moles of X react with 9 moles of Y to form 4 moles of Z.

The correct Option is "4 moles of Z"

A mixture containing 100 g ${{\rm{H}}_2}$ and 100 g ${{\rm{O}}_2}$ is ignited so that water is formed according to the reaction, $2{{\rm{H}}_2} + {{\rm{O}}_2} \to 2{{\rm{H}}_2}{\rm{O}};$ How much water will be formed?

${{\rm{H}}_2} + \frac{1}{2}{{\rm{O}}_2} \to {{\rm{H}}_2}{\rm{O}}$ [Now mole ratio for ${{\rm{H}}_2}:{{\rm{O}}_2}:{{\rm{H}}_2}{\rm{O}}:1:1/2:1;$ Also, ${{\rm{O}}_2}$ is limiting reagent thus] ∴ wt. of ${{\rm{H}}_2}{\rm{O}}$ formed $ = \frac{{100 \times 2}}{{32}} \times 18 = {\rm{ }}112.5{\rm{ }}\;g$

The numerical value of $\frac{{\rm{N}}}{n}$ (where, Nis the number of molecules in a given sample of gas and n is the number of moles of the gas) is

Number of molecules in n moles of substance $ = n \times {N_0}$ $ = n \times 6.023 \times {10^{23}}$ $\frac{{N\left( {{\rm{no}}{\rm{.of}}\;{\rm{ molecules}}} \right)}}{{n\left( {{\rm{no}}{\rm{.of}}\;{\rm{ moles}}} \right)}} = ?$ $ = \frac{{n \times 6.023 \times {{10}^{23}}}}{n} = 6.023 \times {10^{23}}$