(C) We know that $1 \, m = 100 \, cm$ and $1 \, m = 10 \, dm$.Therefore,$100 \, cm = 10 \, dm$.This implies $1 \, cm = \frac{10}{100} \, dm = 0.1 \, d

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(C) We know that $1 \, m = 100 \, cm$ and $1 \, m = 10 \, dm$.Therefore,$100 \, cm = 10 \, dm$.This implies $1 \, cm = \frac{10}{100} \, dm = 0.1 \, dm = 10^{-1} \, dm$.Now,convert $5 \times 10^2 \, cm$ to $dm$:$5 \times 10^2 \, cm = 5 \times 10^2 \times 10^{-1} \, dm = 5 \times 10^1 \, dm$.

Class 11 Chemistry Some Basic Concepts of Chemistry Significant figures and Units for measurement

Easy

$5 \times 10^2 \, cm = .......... \, dm$

A
$5 \times 10^2$
B
$10^1$
C
$5 \times 10^1$
D
$10^2$

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Class 11

Chemistry Questions

Chapter

Some Basic Concepts of Chemistry

Subtopic

Significant figures and Units for measurement

Define Mass and Weight.

Difficult

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Match the following physical quantities with their appropriate units.

Physical Quantity Unit

$(A)$ Molarity $(1)$ $g \ mL^{-1}$

$(B)$ Mole fraction $(2)$ $mol$

$(C)$ Mole $(3)$ $Pascal$

$(D)$ Molality $(4)$ $Unitless$

$(E)$ Pressure $(5)$ $mol \ L^{-1}$

$(F)$ Luminous intensity $(6)$ $Candela$

$(G)$ Density $(7)$ $mol \ kg^{-1}$

$(H)$ Mass $(8)$ $N \ m^{-1}$

$(9)$ $kg$

Physical Quantity	Unit
$(A)$ Molarity	$(1)$ $g \ mL^{-1}$
$(B)$ Mole fraction	$(2)$ $mol$
$(C)$ Mole	$(3)$ $Pascal$
$(D)$ Molality	$(4)$ $Unitless$
$(E)$ Pressure	$(5)$ $mol \ L^{-1}$
$(F)$ Luminous intensity	$(6)$ $Candela$
$(G)$ Density	$(7)$ $mol \ kg^{-1}$
$(H)$ Mass	$(8)$ $N \ m^{-1}$
	$(9)$ $kg$