Tag: properties of substances

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Questions Related to properties of substances

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 is mixed with 30 cc of water. The density of the resulting solution will be

  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

Density of Brine$=\rho _{b}=1.2g/cc$

Volume of Brine$=v _{b}=40cc$
Mass of Brine$=m _{b}=\rho _{b}\times v _{b}=1.2\times 40=48g$
Density of Water$=\rho _{w}=1g/cc$
Volume of Water$=v _{w}=30cc$

Mass of Water$=m _{w}=\rho _{w}\times v _{w}=1\times 30=30g$
Density of mixture$=\dfrac{\text{Mass of mixture}}{\text{ Volume of mixture}}=\dfrac{m _{b}+m _{w}}{v _{b}+v _{w}}=\dfrac{48+30}{40+30}=\dfrac{78}{70}=1.11g/cc$

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

Calculate the mass of air enclosed in a room of length, breadth and height equal to $5\ m, 3 \ m$ and $4 \ m$ respectively. Density of air is $1.3 \ kg/m^3$.

  1. $78 \ kg$
  2. $38 \ kg$
  3. $42 \ kg$
  4. $87 \ kg$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Volume of the room is given as $V = L\times B\times H= 5\times 3\times 4 = 60  { m }^{ 3 }.$
We know, Mass $=$ Density$\times$Volume.
Given that the density $=1.3  kg/m^3$.
So, Mass of the air $=1.3\times60  Kg$.
Hence, mass of the air enclosed in the room is 78 kg.

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