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NEET
Exam
NEET
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

NEET · Chemistry · Some Basic Concepts of Chemistry

Match the column. ( Column I Column II (A) array l No. of g in 20...

Match the column. (
Column IColumn II
(A)No.ofgin20</td></tr><tr><td>molH2SO4\begin{array}{l} No. of g in 20</td></tr><tr><td>mol H_2 SO_4 \end{array}(P)20 N_{A}
(B)No⋅ofmolof</td></tr><tr><td>H2Oin540g\begin{array}{l} No \cdot of mol of</td></tr><tr><td>H_2 O in 540 g \end{array}(Q)6.02 \times 10^{13}
(C)No⋅ofatomsin</td></tr><tr><td>10−10gH2\begin{array}{l} No \cdot of atoms in</td></tr><tr><td>10^{-10} g H_2 \end{array}</td><td>(R)</td><td>30</td></tr><tr><td>(D)</td><td></td><td>(R)</td><td>30</td></tr><tr><td>(D)</td><td>No⋅ofamuin</td></tr><tr><td>20g\begin{array}{l} No \cdot of amu in</td></tr><tr><td>20 g \end{array}</td><td>(S)</td><td>1960</td></tr></table>)</td><td>(S)</td><td>1960</td></tr></table>)

Options

  1. A(A) (\rightarrow) (S), (B) (\rightarrow) (R), (C) (\rightarrow) (Q), (D) (\rightarrow) (P)
  2. B((\mathrm{A}) \rightarrow(\mathrm{P}),(\mathrm{B}) \rightarrow(\mathrm{R}),(\mathrm{C}) \rightarrow(\mathrm{S}),(\mathrm{D}) \rightarrow(\mathrm{Q}))
  3. C((\mathrm{A}) \rightarrow(\mathrm{R}),(\mathrm{B}) \rightarrow(\mathrm{S}),(\mathrm{C}) \rightarrow(\mathrm{P}),(\mathrm{D}) \rightarrow(\mathrm{Q}))
  4. D(A) (\rightarrow) (P), (B) (\rightarrow) (R), (C) (\rightarrow) (Q), (D) (\rightarrow) (S)

Correct answer

A. (A) (\rightarrow) (S), (B) (\rightarrow) (R), (C) (\rightarrow) (Q), (D) (\rightarrow) (P)

Step-by-step solution

(A) No. of gg in 20 mol (\mathrm{H}_2 \mathrm{SO}_4) (\mathrm{H}_2 \mathrm{SO}_4=) Molar mass of (\text{ Mass }=20, \text{ mol } \times 98) - Matches with (S). (B) No. of mol of (\mathrm{H}_2 \mathrm{O}) in (\mathrm{5 4 0 g}) (\mathrm{H}_2 \mathrm{O}=2) Molar mass of - Moles (=\frac{540, \mathrm{ g}}{18, \mathrm{ g} / \mathrm{mol}}=) - Matches with (R). (C) No. of atoms in (10^{-10}, \mathrm{ g}, \mathrm{H}_2) - Molar mass of (\mathrm{H}_2=\mathrm{2}) Moles =10−10,g2,g/mol=55×10−11,mol×6.0\begin{aligned} \text{Moles }=\frac{\frac{10^{-10}, \mathrm{ g}}{2, \mathrm{ g} / \mathrm{mol}}}{=} & 5 \\ 5 & \times 10^{-11}, \mathrm{ mol} \times 6.0 \end{aligned}) No.ofofatoms = atoms - Matches with (Q). (D) No. of amu in (\mathrm{2 0 g}) (\cdot 1 \mathrm{ g}=1000 \mathrm{mg}), hence 20 g corresponds to 20 amu when converted directly. - Matches with (P). Summary of Matches - ((\mathrm{A}) \rightarrow(\mathrm{S})) - (B) (\rightarrow) (R) - (C) (\rightarrow) (Q) - (D) (\rightarrow) (P)$
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