Tag: substances in the surroundings - their states and properties

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

The density of a cuboid of mass 200 g with dimensions 2 cm $\times$ 4 cm $\times$ 5 cm is

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

  1. 1000 kg m$^{-3}$

  2. 3000 kg m$^{-3}$

  3. 5000 kg m$^{-3}$

  4. 2000 kg m$^{-3}$

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Correct Option: C
Explanation:
The density of a body is defined as the amount of substance contained per unit volume.
Density $=\dfrac{mass}{volume}$

$\displaystyle d = \frac{0.2 kg}{40  cm^3} = \frac{0.2 \times 10^6}{40} kg  m^{-3}$

$=5000  kg  m^{-3}$

A wire of length 50 cm has a mass of 20 gm. It its radius is halved by stretching, its new mass per unit length will be

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

  1. 0.4 g cm$^{-1}$

  2. 0.2 kg m$^{-1}$

  3. 0.1 g cm$^{-1}$

  4. 0.2 g cm$^{-1}$

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Correct Option: C
Explanation:

On stretching, volume remains constant. Hence on decreasing the radius by half, the length of the wire becomes quadrupled. 
Mass per unit length $\displaystyle = \frac{20}{4 \times 50 cm}$
$=0.1  g  cm^{-1}$

Density of water is :

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

  1. 1 g cm$^{-3}$

  2. 100 kg m$^{-3}$

  3. 1000 kg m$^{-3}$

  4. None

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Correct Option: A,C
Explanation:

 Density of water = 1 g $ cm^{-3 }$ = 1000 kg $ m^{-3}$

so option (A) and option (C) are correct

If 5 litres of kerosene has a mass of 5 kg, then what is the density of kerosene?

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

  1. 500 kg/$m^3$

  2. 1000 kg/$m^3$

  3. 100 kg/$m^3$

  4. 50 kg/$m^3$

Show answer
Correct Option: B
Explanation:

Density of kerosene $= \displaystyle \frac{mass}{volume}$
$\displaystyle = \frac{5  kg}{5  litres}= \frac{5kg}{5 \times 10^{-3}m^3} = 1000 kg  m^{-3}$

Two spheres are having the ratio of their densities 1:3 and masses in 3:4 respectively. Find the ratio of their volumes. i.e $\dfrac{V _1}{V _2}$

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

  1. 1:3

  2. 3:1

  3. 4:9

  4. 9:4

Show answer
Correct Option: D
Explanation:

$\displaystyle \frac{d _1}{d _2} = \frac{1}{3} ; \frac{m _1}{m _2} = \frac{3}{4}$
$d = m/v$
$\therefore \displaystyle \frac{V _1}{V _2}  = \frac{m _1}{m _2} \times \frac{d _2}{d _1} = \frac{3}{4} \times \frac{3}{1} = 9:4$

Density of air at room temperature is

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

  1. $120 \ {g\ cm}^{-3}$

  2. $120 \ {g\ mm}^{-3}$

  3. $12 \ {g\ m}^{-3}$

  4. $1.3 \ {kg\ m}^{-3}$

Show answer
Correct Option: D
Explanation:

The density of air at NTP is $1.293 kg m^{-3}$.

So after rounding, it is $1.3 kg m^{-3}$.

If the mass of a body is 12.1 g and its density is 2.2 g/cc, its volume is.

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

  1. $5.5 cm^3$

  2. $8$ cc

  3. $11$ cc

  4. $55$ cc

Show answer
Correct Option: A
Explanation:

Given:

mass $(m)=12.1g$
density $(d)=2.2g/cc$
Let the volume be $v$
As $density=\dfrac{mass}{volume}$
$d=\dfrac{m}{v}$
$v=\dfrac{m}{d}$
$v=\dfrac{12.1}{2.2}=5.5cc$

The density of water is equal to :

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

  1. $10^{-3}$ $kgm^{-3}$

  2. $10^{-2}$ $kgm^{-3}$

  3. $10^{2}$ $kgm^{-3}$

  4. $10^{3}$ $kgm^{-3}$

Show answer
Correct Option: D
Explanation:

density of water in S.I. unit= ${10}^3$ kg/${m}^3$

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

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

  1. $2.11$ g/cc

  2. $1.11$ g/cc

  3. $12.2$ g/cc

  4. $20.4$ g/cc

Show answer
Correct Option: B
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$=\dfrac{\text{Mass of mixture}}{\text{ Volume of mixture}}=\dfrac{m _{b}+m _{w}}{v _{b}+v _{w}}=\dfrac{48+30}{40+30}=\dfrac{78}{70}=1.11g/cc$

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 $1.3 \ kg/m^3$.

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

  1. $78 \ kg$

  2. $38 \ kg$

  3. $42 \  kg$

  4. $87 \ kg$

Show answer
Correct Option: A
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.