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CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306809 If we consider that $\frac{1}{6}$ , in place of $\frac{1}{12}$ , mass of carbon atom is taken to be the relative mass unit, the mass of one mole of a substance will

1 Decrease twice

2 Increase two fold

3 Remain unchanged

4 Be a function of the molecular mass of the substance

Explanation:

Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{\frac{1}{1 2^{t h}} p a r t o f t h e m a s s o f o n e a t o m o f C a r b o n - 12}$
$o r \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{m a s s o f o n e a t o m o f t h e C - 12} \times 12$
Now if we use $\frac{1}{6}$ in place $\frac{1}{12}$ the formula becomes Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f e l e m e n t}{M a s s o f o n e a t o m o f C - 12} \times 6$
$∴$ Relative atomic mass decrease twice

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules $(d = 0.75 g L^{- 1}), 11.2 g$ of molecules $(d = 3 g L^{- 1})$ and 5.6 g of molecules $(d = 1 . 5 g L^{- 1})$ . All density measurements are made at STP. Calculate the total number of molecules ( $N$ ) present in the given sample. Report your answer in ' $10^{23} N$ '. Assume Avogadro's number as $6 \times 10^{23}$ .

1 0.5

2 5

3 3

4 0.05

Explanation:

Total weight $= 19.6 g$
(a) 2.8 g of molecules, $d = 0.765 g L^{- 1}$
$V o l u m e = \frac{M a s s}{d} = \frac{2 .8}{0 .75} L a t S T P$
$1 m o l \equiv 22.4 L a t S T P$
$m o l e s = \frac{2.8}{0.75} \times \frac{1}{22.4}$
$M o l e c u l e s = \frac{2.8}{0.75} \times \frac{1}{22.4} \times 6 \times 10^{23}$
$= 1 \times 10^{23}$
(b) 11.2 g molecules, $d = 3 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
(c) 5.6 g molecules, $d = 1.5 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
Total number of molecules $= 3 \times 10^{23} = 3$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306789 An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
$= 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

6 .023 \times 1 0^{9}

2

6 .023 \times 1 0^{23}

3

6 .023 \times 1 0^{21}

4

6 .023 \times 1 0^{20}

Explanation:

$C_{x} H_{y} C l$ is the formula of chorohydrocarbon.
Given, % Cl = 3.55%
Weight of $C l = 1 \times \frac{3 .55}{100}$
$n_{C l^{- 1}} = \frac{1 \times 3 .55}{100 \times 35 .5}$
No of $C l^{- 1} i o n = \frac{1 \times 3 .55}{100 \times 35 .5} \times 6 .023 \times 1 0^{23}$
$= 6 .023 \times 1 0^{20}$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306790 The number of moles of $H_{2} O$ in one litre is

1 50.5

2 55

3 55.05

4 55.55

Explanation:

Density of water $= 1 g / cc$
Mass of water $=$ Density $\times$ Volume $= 1 \times 1000 = 1000 g$
$[∵ 1 L = 1000 cc]$
$N o . o f m o l e s o f H_{2} O = \frac{M a s s o f H_{2} O}{M o l a r m a s s o f H_{2} O}$
$= \frac{1000}{18} = 55.55$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306809 If we consider that $\frac{1}{6}$ , in place of $\frac{1}{12}$ , mass of carbon atom is taken to be the relative mass unit, the mass of one mole of a substance will

1 Decrease twice

2 Increase two fold

3 Remain unchanged

4 Be a function of the molecular mass of the substance

Explanation:

Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{\frac{1}{1 2^{t h}} p a r t o f t h e m a s s o f o n e a t o m o f C a r b o n - 12}$
$o r \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{m a s s o f o n e a t o m o f t h e C - 12} \times 12$
Now if we use $\frac{1}{6}$ in place $\frac{1}{12}$ the formula becomes Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f e l e m e n t}{M a s s o f o n e a t o m o f C - 12} \times 6$
$∴$ Relative atomic mass decrease twice

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules $(d = 0.75 g L^{- 1}), 11.2 g$ of molecules $(d = 3 g L^{- 1})$ and 5.6 g of molecules $(d = 1 . 5 g L^{- 1})$ . All density measurements are made at STP. Calculate the total number of molecules ( $N$ ) present in the given sample. Report your answer in ' $10^{23} N$ '. Assume Avogadro's number as $6 \times 10^{23}$ .

1 0.5

2 5

3 3

4 0.05

Explanation:

Total weight $= 19.6 g$
(a) 2.8 g of molecules, $d = 0.765 g L^{- 1}$
$V o l u m e = \frac{M a s s}{d} = \frac{2 .8}{0 .75} L a t S T P$
$1 m o l \equiv 22.4 L a t S T P$
$m o l e s = \frac{2.8}{0.75} \times \frac{1}{22.4}$
$M o l e c u l e s = \frac{2.8}{0.75} \times \frac{1}{22.4} \times 6 \times 10^{23}$
$= 1 \times 10^{23}$
(b) 11.2 g molecules, $d = 3 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
(c) 5.6 g molecules, $d = 1.5 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
Total number of molecules $= 3 \times 10^{23} = 3$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306789 An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
$= 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

6 .023 \times 1 0^{9}

2

6 .023 \times 1 0^{23}

3

6 .023 \times 1 0^{21}

4

6 .023 \times 1 0^{20}

Explanation:

$C_{x} H_{y} C l$ is the formula of chorohydrocarbon.
Given, % Cl = 3.55%
Weight of $C l = 1 \times \frac{3 .55}{100}$
$n_{C l^{- 1}} = \frac{1 \times 3 .55}{100 \times 35 .5}$
No of $C l^{- 1} i o n = \frac{1 \times 3 .55}{100 \times 35 .5} \times 6 .023 \times 1 0^{23}$
$= 6 .023 \times 1 0^{20}$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306790 The number of moles of $H_{2} O$ in one litre is

1 50.5

2 55

3 55.05

4 55.55

Explanation:

Density of water $= 1 g / cc$
Mass of water $=$ Density $\times$ Volume $= 1 \times 1000 = 1000 g$
$[∵ 1 L = 1000 cc]$
$N o . o f m o l e s o f H_{2} O = \frac{M a s s o f H_{2} O}{M o l a r m a s s o f H_{2} O}$
$= \frac{1000}{18} = 55.55$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306809 If we consider that $\frac{1}{6}$ , in place of $\frac{1}{12}$ , mass of carbon atom is taken to be the relative mass unit, the mass of one mole of a substance will

1 Decrease twice

2 Increase two fold

3 Remain unchanged

4 Be a function of the molecular mass of the substance

Explanation:

Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{\frac{1}{1 2^{t h}} p a r t o f t h e m a s s o f o n e a t o m o f C a r b o n - 12}$
$o r \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{m a s s o f o n e a t o m o f t h e C - 12} \times 12$
Now if we use $\frac{1}{6}$ in place $\frac{1}{12}$ the formula becomes Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f e l e m e n t}{M a s s o f o n e a t o m o f C - 12} \times 6$
$∴$ Relative atomic mass decrease twice

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules $(d = 0.75 g L^{- 1}), 11.2 g$ of molecules $(d = 3 g L^{- 1})$ and 5.6 g of molecules $(d = 1 . 5 g L^{- 1})$ . All density measurements are made at STP. Calculate the total number of molecules ( $N$ ) present in the given sample. Report your answer in ' $10^{23} N$ '. Assume Avogadro's number as $6 \times 10^{23}$ .

1 0.5

2 5

3 3

4 0.05

Explanation:

Total weight $= 19.6 g$
(a) 2.8 g of molecules, $d = 0.765 g L^{- 1}$
$V o l u m e = \frac{M a s s}{d} = \frac{2 .8}{0 .75} L a t S T P$
$1 m o l \equiv 22.4 L a t S T P$
$m o l e s = \frac{2.8}{0.75} \times \frac{1}{22.4}$
$M o l e c u l e s = \frac{2.8}{0.75} \times \frac{1}{22.4} \times 6 \times 10^{23}$
$= 1 \times 10^{23}$
(b) 11.2 g molecules, $d = 3 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
(c) 5.6 g molecules, $d = 1.5 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
Total number of molecules $= 3 \times 10^{23} = 3$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306789 An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
$= 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

6 .023 \times 1 0^{9}

2

6 .023 \times 1 0^{23}

3

6 .023 \times 1 0^{21}

4

6 .023 \times 1 0^{20}

Explanation:

$C_{x} H_{y} C l$ is the formula of chorohydrocarbon.
Given, % Cl = 3.55%
Weight of $C l = 1 \times \frac{3 .55}{100}$
$n_{C l^{- 1}} = \frac{1 \times 3 .55}{100 \times 35 .5}$
No of $C l^{- 1} i o n = \frac{1 \times 3 .55}{100 \times 35 .5} \times 6 .023 \times 1 0^{23}$
$= 6 .023 \times 1 0^{20}$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306790 The number of moles of $H_{2} O$ in one litre is

1 50.5

2 55

3 55.05

4 55.55

Explanation:

Density of water $= 1 g / cc$
Mass of water $=$ Density $\times$ Volume $= 1 \times 1000 = 1000 g$
$[∵ 1 L = 1000 cc]$
$N o . o f m o l e s o f H_{2} O = \frac{M a s s o f H_{2} O}{M o l a r m a s s o f H_{2} O}$
$= \frac{1000}{18} = 55.55$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306809 If we consider that $\frac{1}{6}$ , in place of $\frac{1}{12}$ , mass of carbon atom is taken to be the relative mass unit, the mass of one mole of a substance will

1 Decrease twice

2 Increase two fold

3 Remain unchanged

4 Be a function of the molecular mass of the substance

Explanation:

Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{\frac{1}{1 2^{t h}} p a r t o f t h e m a s s o f o n e a t o m o f C a r b o n - 12}$
$o r \frac{M a s s o f o n e a t o m o f t h e e l e m e n t}{m a s s o f o n e a t o m o f t h e C - 12} \times 12$
Now if we use $\frac{1}{6}$ in place $\frac{1}{12}$ the formula becomes Relative atomic mass
$= \frac{M a s s o f o n e a t o m o f e l e m e n t}{M a s s o f o n e a t o m o f C - 12} \times 6$
$∴$ Relative atomic mass decrease twice

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306788 A 19.6 g of a given gaseous sample contains 2.8 g of molecules $(d = 0.75 g L^{- 1}), 11.2 g$ of molecules $(d = 3 g L^{- 1})$ and 5.6 g of molecules $(d = 1 . 5 g L^{- 1})$ . All density measurements are made at STP. Calculate the total number of molecules ( $N$ ) present in the given sample. Report your answer in ' $10^{23} N$ '. Assume Avogadro's number as $6 \times 10^{23}$ .

1 0.5

2 5

3 3

4 0.05

Explanation:

Total weight $= 19.6 g$
(a) 2.8 g of molecules, $d = 0.765 g L^{- 1}$
$V o l u m e = \frac{M a s s}{d} = \frac{2 .8}{0 .75} L a t S T P$
$1 m o l \equiv 22.4 L a t S T P$
$m o l e s = \frac{2.8}{0.75} \times \frac{1}{22.4}$
$M o l e c u l e s = \frac{2.8}{0.75} \times \frac{1}{22.4} \times 6 \times 10^{23}$
$= 1 \times 10^{23}$
(b) 11.2 g molecules, $d = 3 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
(c) 5.6 g molecules, $d = 1.5 g L^{- 1}$
Molecules $= 1 \times 10^{23}$
Total number of molecules $= 3 \times 10^{23} = 3$

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306789 An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl u = 35.5u ; Avogadro number
$= 6 .023 \times 1 0^{23} m o l^{- 1}$ )

1

6 .023 \times 1 0^{9}

2

6 .023 \times 1 0^{23}

3

6 .023 \times 1 0^{21}

4

6 .023 \times 1 0^{20}

Explanation:

$C_{x} H_{y} C l$ is the formula of chorohydrocarbon.
Given, % Cl = 3.55%
Weight of $C l = 1 \times \frac{3 .55}{100}$
$n_{C l^{- 1}} = \frac{1 \times 3 .55}{100 \times 35 .5}$
No of $C l^{- 1} i o n = \frac{1 \times 3 .55}{100 \times 35 .5} \times 6 .023 \times 1 0^{23}$
$= 6 .023 \times 1 0^{20}$

JEE - 2018

CHXI01:SOME BASIC CONCEPTS OF CHEMISTRY

306790 The number of moles of $H_{2} O$ in one litre is

1 50.5

2 55

3 55.05

4 55.55

Explanation:

Density of water $= 1 g / cc$
Mass of water $=$ Density $\times$ Volume $= 1 \times 1000 = 1000 g$
$[∵ 1 L = 1000 cc]$
$N o . o f m o l e s o f H_{2} O = \frac{M a s s o f H_{2} O}{M o l a r m a s s o f H_{2} O}$
$= \frac{1000}{18} = 55.55$