Tag: substances in the surroundings - their states and properties

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Questions Related to substances in the surroundings - their states and properties

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

If two masses A and B have their masses in the ratio 1 : 4 and their volumes are equal, then their densities have the ratio

  1. 1:4

  2. 8:1

  3. 2: 4

  4. 3:1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let mass of $A$ be $m$

Mass of $B$ be $4m$
Let density of $A$ be $d _a$
Density of $B$ be $d _b$
And Volume of A$=$Volume of B$=V$
As Density$=\dfrac{\text {Mass}}{\text {Volume}}$

Therefore $d _a=\dfrac{m}{V}$
$d _b=\dfrac{4m}{V}$

$\dfrac{d _a}{d _b}=\dfrac{\dfrac{m}{V}}{\dfrac{4m}{V}}$
$\dfrac{d _a}{d _b}=\dfrac{m}{4m}=\dfrac{1}{4}$
Hence the correct answer is option (A).

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

What is the density of an iron block which has a volume of ${12 cm^3}$ and a mass of 96g?

  1. ${ 8 kg/m ^3}$
  2. ${80 kg /m ^3}$
  3. ${8000 kg /m ^3}$
  4. ${800 kg/ m ^3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

As we know, Density $d = \dfrac{Mass, M}{ Volume, V}$


In this case, an iron block which has a volume of ${12 cm^3}$ = ${0.12 m^3}$ and a mass of $96 g = 0.096 kg$.

Therefore, Density $d =\dfrac{ 0.096}{0.12} = 8 kg/m^{ 3 }$.

Hence, the density of the iron block is 8 $kg/m^{ 3 }$.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The ratio of the values in SI units to values in CGS units of density is

  1. $10^3:1$
  2. $10^2:1$
  3. $10^{-2}:1$
  4. $10^{-3}:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

the density of water in S.I. unit $\rho=1000 kg/{m}^3$

the density of water in CGS unit $\rho'=1 gm/cc$

the ratio of the densities of water in the different system 
$\dfrac{\rho}{\rho'}=\dfrac{{10}^3}{1}$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

When air is cooled, its density increases. State whether true or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 As on cooling volume of air decreases and its mass remains constant so its density increases as  Density = $\dfrac{mass}{volume}$

hence given statement is correct so option (A) is correct

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The hot air balloon rises because it is

  1. denser

  2. less dense

  3. equally dense

  4. the given statement is wrong
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The hot air balloon has the hot air filled in it and the density of air is more than the hot air. So, the normal air or atmospheric air has the tendency to remain below than the hot air as the objects having the more density remains below than the objects having less density.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Brine has a density of $1.2  {g}/{cc}$. $40  cc$ of it are mixed with $30  cc$ of water. The density of solution is

  1. $2.11 {g}/{cc}$
  2. $1.11 {g}/{cc}$
  3. $12.2 {g}/{cc}$
  4. $20.4 {g}/{cc}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
given,  ${ \rho } _{ a }=1.2\quad 9/cc\quad \quad \quad { V } _{ a }=40cc$
                 ${ \rho } _{ b }=1g/cc\quad \quad \quad { V } _{ b }=30cc$

$ \rho _{mixture} = \dfrac { { \rho } _{ a }{ V } _{ a }+{ \rho } _{ b }{ V } _{ b } }{ { V } _{ a }+{ V } _{ b } } $

So  $ \rho _{mixture} = \dfrac { 1.2\times 40+1\times 30 }{ 40+30 } =\dfrac { 78 }{ 70 } $

        $\boxed { \rho _{mixture}=1.11\quad g/cc } $
Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

With the increase in temperature, the density of a substance, in general, ____________.

  1. increases

  2. decreases

  3. first increase then decreases

  4. first decrease then increases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Due to increase in temperature, objects vibrational energy increases resulting in decrease in density of the substance.