QBManager

Some Basic Concepts of Chemistry

228748 $1 g$ of silver gets distributed between $10 {cm}^{3}$ of molten zinc and $100 {cm}^{3}$ of molten lead at $800^{\circ} C$ . The percentage of silver in the zinc layer is approximately

1 89

2 91

3 97

4 94

Explanation:

Let mass of $Ag$ in $Zn$ layer $= x$
$∴$ Mass of $Ag$ in lead layer $= 1 - x$
Concentration of Ag is zinc $= \frac{x}{10}$
$∴$ Concentration of $Ag$ in $Pb = \frac{1 - x}{100}$
$\frac{\frac{x}{10}}{\frac{1 - x}{100}} = 300$
$% v$ of $Ag$ in Zinc layer $= 97 %$

BCECE-2014

Some Basic Concepts of Chemistry

228749 In Dumas' method of estimation of nitrogen $0.35 g$ of an organic compound gave $55 mL$ . of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at $300 K = 15 mm$ )

1 15.44

2 16.45

3 17.45

4 14.45

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Given,
Where, $p_{2} =$ pressure of $N_{2}$ at $STP = 760 mm$
$T_{2} =$ Temperature of $N_{2}$ at $STP = 273 K$
Volume of $N_{2}$ at STP (by gas equation)
$(\frac{P - P_{1}}{t + 273}) V_{1} \times \frac{273}{160} = V_{2}$
$P = 715 mm$ (pressure at which $N_{2}$ collected)
$P_{1} =$ aqueous tension of water $= 15 mm$
$T_{1} = t + 273 = 300 K$
$V_{1} = 55 ml$
$∴ V_{2} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.098 mL$
$%$ of nitrogen in given compound
$= \frac{28}{22400} \times \frac{V_{2}}{W} \times 100 = \frac{28}{22400} \times \frac{46.09}{0.35} \times 100$
$= 16.45 %$

JIPMER-2014 NEET-2011

Some Basic Concepts of Chemistry

228751 A mixture of ${CaCl}_{2}$ and $NaCl$ weighing $4.44 g$ is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to set $0.56 g$ of $CaO$ . The percentage of $NaCl$ in the mixture is (atomic mass of $Ca = 40$ )

1 75

2 31.5

3 40.2

4 25

Explanation:

Concerned reaction is.
${CaCl}_{2} + NaCl + {Na}_{2} {CO}_{3} \to {CaCO}_{3} + 2 NaCl + NaCl$
${CaCO}_{3} \overset{Δ}{\to} {CaO}_{(0.56 g)} + {CO}_{2}$
$100 {CaCO}_{3} = 111 g {CaCl}_{2}$
$1 g {CaCO}_{3} = 1.11 g {CaCl}_{2}$
$weight of {CaCl}_{2} = 1.11 g, weight of NaCl = 4.44 - 1.11 = 3.33 g$

2010

Some Basic Concepts of Chemistry

228752 An organic compound contains $60 % C; 4.48 %$ $H$ and $35.5 %$ O. Its empirical formula is

1

C_{9} H_{8} O_{4}

2

C_{5} H_{4} O_{2}

3

C_{5} H_{4} O_{4}

4

C_{9} H_{7} O_{6}

Explanation:

Element \% Atomic mass Moles
Carbon $60 12 \frac{60}{12} \Rightarrow 5$
Oxygen $35.5 16 \frac{35.5}{16} \Rightarrow 2.2$
Empirical formula $= C : H : O$
$= \frac{5}{2.2} : \frac{4.48}{2.2} : \frac{2.2}{2.2}$
$= 2.27 : 2.0 : 1.0$
$= 2.27 \times 4 : 2 \times 4 : 1 \times 4$
$= 9 : 8 : 4$
$= C_{9} H_{8} O_{4}$

TS-EAMCET (Engg.)

Some Basic Concepts of Chemistry

228753 If $0.765 g$ of an acid gives $0.535 g$ of ${CO}_{2}$ and $0.138 g$ of $H_{2} O$ then ratio percentage of $C$ and $H$ is

1

19 : 2

2

18 : 12

3

20 : 17

4

1 : 7

Explanation:

Given -
Mass of acid $= 0.765 gm$
Mass of ${CO}_{2} = 0.535 gm$
Mass of $H_{2} O = 0.138 gm$
$∴$ Acid $\times {CO}_{2} + H_{2} O$
$% C = \frac{12}{44} \times \frac{mass of {CO}_{2}}{mass of acid (compound)} \times 100$
$% C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 = 19.07 %$
$\begin{aligned} % H = \frac{2}{18} \times \frac{mass of H_{2} O}{mass of compound} \times 100 \\ % H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 = 2 % \end{aligned}$
So, the ratio of percentage of $C$ and $H$ is $19 : 2$ .

Shift-II

Some Basic Concepts of Chemistry

228748 $1 g$ of silver gets distributed between $10 {cm}^{3}$ of molten zinc and $100 {cm}^{3}$ of molten lead at $800^{\circ} C$ . The percentage of silver in the zinc layer is approximately

1 89

2 91

3 97

4 94

Explanation:

Let mass of $Ag$ in $Zn$ layer $= x$
$∴$ Mass of $Ag$ in lead layer $= 1 - x$
Concentration of Ag is zinc $= \frac{x}{10}$
$∴$ Concentration of $Ag$ in $Pb = \frac{1 - x}{100}$
$\frac{\frac{x}{10}}{\frac{1 - x}{100}} = 300$
$% v$ of $Ag$ in Zinc layer $= 97 %$

BCECE-2014

Some Basic Concepts of Chemistry

228749 In Dumas' method of estimation of nitrogen $0.35 g$ of an organic compound gave $55 mL$ . of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at $300 K = 15 mm$ )

1 15.44

2 16.45

3 17.45

4 14.45

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Given,
Where, $p_{2} =$ pressure of $N_{2}$ at $STP = 760 mm$
$T_{2} =$ Temperature of $N_{2}$ at $STP = 273 K$
Volume of $N_{2}$ at STP (by gas equation)
$(\frac{P - P_{1}}{t + 273}) V_{1} \times \frac{273}{160} = V_{2}$
$P = 715 mm$ (pressure at which $N_{2}$ collected)
$P_{1} =$ aqueous tension of water $= 15 mm$
$T_{1} = t + 273 = 300 K$
$V_{1} = 55 ml$
$∴ V_{2} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.098 mL$
$%$ of nitrogen in given compound
$= \frac{28}{22400} \times \frac{V_{2}}{W} \times 100 = \frac{28}{22400} \times \frac{46.09}{0.35} \times 100$
$= 16.45 %$

JIPMER-2014 NEET-2011

Some Basic Concepts of Chemistry

228751 A mixture of ${CaCl}_{2}$ and $NaCl$ weighing $4.44 g$ is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to set $0.56 g$ of $CaO$ . The percentage of $NaCl$ in the mixture is (atomic mass of $Ca = 40$ )

1 75

2 31.5

3 40.2

4 25

Explanation:

Concerned reaction is.
${CaCl}_{2} + NaCl + {Na}_{2} {CO}_{3} \to {CaCO}_{3} + 2 NaCl + NaCl$
${CaCO}_{3} \overset{Δ}{\to} {CaO}_{(0.56 g)} + {CO}_{2}$
$100 {CaCO}_{3} = 111 g {CaCl}_{2}$
$1 g {CaCO}_{3} = 1.11 g {CaCl}_{2}$
$weight of {CaCl}_{2} = 1.11 g, weight of NaCl = 4.44 - 1.11 = 3.33 g$

2010

Some Basic Concepts of Chemistry

228752 An organic compound contains $60 % C; 4.48 %$ $H$ and $35.5 %$ O. Its empirical formula is

1

C_{9} H_{8} O_{4}

2

C_{5} H_{4} O_{2}

3

C_{5} H_{4} O_{4}

4

C_{9} H_{7} O_{6}

Explanation:

Element \% Atomic mass Moles
Carbon $60 12 \frac{60}{12} \Rightarrow 5$
Oxygen $35.5 16 \frac{35.5}{16} \Rightarrow 2.2$
Empirical formula $= C : H : O$
$= \frac{5}{2.2} : \frac{4.48}{2.2} : \frac{2.2}{2.2}$
$= 2.27 : 2.0 : 1.0$
$= 2.27 \times 4 : 2 \times 4 : 1 \times 4$
$= 9 : 8 : 4$
$= C_{9} H_{8} O_{4}$

TS-EAMCET (Engg.)

Some Basic Concepts of Chemistry

228753 If $0.765 g$ of an acid gives $0.535 g$ of ${CO}_{2}$ and $0.138 g$ of $H_{2} O$ then ratio percentage of $C$ and $H$ is

1

19 : 2

2

18 : 12

3

20 : 17

4

1 : 7

Explanation:

Given -
Mass of acid $= 0.765 gm$
Mass of ${CO}_{2} = 0.535 gm$
Mass of $H_{2} O = 0.138 gm$
$∴$ Acid $\times {CO}_{2} + H_{2} O$
$% C = \frac{12}{44} \times \frac{mass of {CO}_{2}}{mass of acid (compound)} \times 100$
$% C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 = 19.07 %$
$\begin{aligned} % H = \frac{2}{18} \times \frac{mass of H_{2} O}{mass of compound} \times 100 \\ % H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 = 2 % \end{aligned}$
So, the ratio of percentage of $C$ and $H$ is $19 : 2$ .

Shift-II

Some Basic Concepts of Chemistry

228748 $1 g$ of silver gets distributed between $10 {cm}^{3}$ of molten zinc and $100 {cm}^{3}$ of molten lead at $800^{\circ} C$ . The percentage of silver in the zinc layer is approximately

1 89

2 91

3 97

4 94

Explanation:

Let mass of $Ag$ in $Zn$ layer $= x$
$∴$ Mass of $Ag$ in lead layer $= 1 - x$
Concentration of Ag is zinc $= \frac{x}{10}$
$∴$ Concentration of $Ag$ in $Pb = \frac{1 - x}{100}$
$\frac{\frac{x}{10}}{\frac{1 - x}{100}} = 300$
$% v$ of $Ag$ in Zinc layer $= 97 %$

BCECE-2014

Some Basic Concepts of Chemistry

228749 In Dumas' method of estimation of nitrogen $0.35 g$ of an organic compound gave $55 mL$ . of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at $300 K = 15 mm$ )

1 15.44

2 16.45

3 17.45

4 14.45

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Given,
Where, $p_{2} =$ pressure of $N_{2}$ at $STP = 760 mm$
$T_{2} =$ Temperature of $N_{2}$ at $STP = 273 K$
Volume of $N_{2}$ at STP (by gas equation)
$(\frac{P - P_{1}}{t + 273}) V_{1} \times \frac{273}{160} = V_{2}$
$P = 715 mm$ (pressure at which $N_{2}$ collected)
$P_{1} =$ aqueous tension of water $= 15 mm$
$T_{1} = t + 273 = 300 K$
$V_{1} = 55 ml$
$∴ V_{2} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.098 mL$
$%$ of nitrogen in given compound
$= \frac{28}{22400} \times \frac{V_{2}}{W} \times 100 = \frac{28}{22400} \times \frac{46.09}{0.35} \times 100$
$= 16.45 %$

JIPMER-2014 NEET-2011

Some Basic Concepts of Chemistry

228751 A mixture of ${CaCl}_{2}$ and $NaCl$ weighing $4.44 g$ is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to set $0.56 g$ of $CaO$ . The percentage of $NaCl$ in the mixture is (atomic mass of $Ca = 40$ )

1 75

2 31.5

3 40.2

4 25

Explanation:

Concerned reaction is.
${CaCl}_{2} + NaCl + {Na}_{2} {CO}_{3} \to {CaCO}_{3} + 2 NaCl + NaCl$
${CaCO}_{3} \overset{Δ}{\to} {CaO}_{(0.56 g)} + {CO}_{2}$
$100 {CaCO}_{3} = 111 g {CaCl}_{2}$
$1 g {CaCO}_{3} = 1.11 g {CaCl}_{2}$
$weight of {CaCl}_{2} = 1.11 g, weight of NaCl = 4.44 - 1.11 = 3.33 g$

2010

Some Basic Concepts of Chemistry

228752 An organic compound contains $60 % C; 4.48 %$ $H$ and $35.5 %$ O. Its empirical formula is

1

C_{9} H_{8} O_{4}

2

C_{5} H_{4} O_{2}

3

C_{5} H_{4} O_{4}

4

C_{9} H_{7} O_{6}

Explanation:

Element \% Atomic mass Moles
Carbon $60 12 \frac{60}{12} \Rightarrow 5$
Oxygen $35.5 16 \frac{35.5}{16} \Rightarrow 2.2$
Empirical formula $= C : H : O$
$= \frac{5}{2.2} : \frac{4.48}{2.2} : \frac{2.2}{2.2}$
$= 2.27 : 2.0 : 1.0$
$= 2.27 \times 4 : 2 \times 4 : 1 \times 4$
$= 9 : 8 : 4$
$= C_{9} H_{8} O_{4}$

TS-EAMCET (Engg.)

Some Basic Concepts of Chemistry

228753 If $0.765 g$ of an acid gives $0.535 g$ of ${CO}_{2}$ and $0.138 g$ of $H_{2} O$ then ratio percentage of $C$ and $H$ is

1

19 : 2

2

18 : 12

3

20 : 17

4

1 : 7

Explanation:

Given -
Mass of acid $= 0.765 gm$
Mass of ${CO}_{2} = 0.535 gm$
Mass of $H_{2} O = 0.138 gm$
$∴$ Acid $\times {CO}_{2} + H_{2} O$
$% C = \frac{12}{44} \times \frac{mass of {CO}_{2}}{mass of acid (compound)} \times 100$
$% C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 = 19.07 %$
$\begin{aligned} % H = \frac{2}{18} \times \frac{mass of H_{2} O}{mass of compound} \times 100 \\ % H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 = 2 % \end{aligned}$
So, the ratio of percentage of $C$ and $H$ is $19 : 2$ .

Shift-II

Some Basic Concepts of Chemistry

228748 $1 g$ of silver gets distributed between $10 {cm}^{3}$ of molten zinc and $100 {cm}^{3}$ of molten lead at $800^{\circ} C$ . The percentage of silver in the zinc layer is approximately

1 89

2 91

3 97

4 94

Explanation:

Let mass of $Ag$ in $Zn$ layer $= x$
$∴$ Mass of $Ag$ in lead layer $= 1 - x$
Concentration of Ag is zinc $= \frac{x}{10}$
$∴$ Concentration of $Ag$ in $Pb = \frac{1 - x}{100}$
$\frac{\frac{x}{10}}{\frac{1 - x}{100}} = 300$
$% v$ of $Ag$ in Zinc layer $= 97 %$

BCECE-2014

Some Basic Concepts of Chemistry

228749 In Dumas' method of estimation of nitrogen $0.35 g$ of an organic compound gave $55 mL$ . of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at $300 K = 15 mm$ )

1 15.44

2 16.45

3 17.45

4 14.45

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Given,
Where, $p_{2} =$ pressure of $N_{2}$ at $STP = 760 mm$
$T_{2} =$ Temperature of $N_{2}$ at $STP = 273 K$
Volume of $N_{2}$ at STP (by gas equation)
$(\frac{P - P_{1}}{t + 273}) V_{1} \times \frac{273}{160} = V_{2}$
$P = 715 mm$ (pressure at which $N_{2}$ collected)
$P_{1} =$ aqueous tension of water $= 15 mm$
$T_{1} = t + 273 = 300 K$
$V_{1} = 55 ml$
$∴ V_{2} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.098 mL$
$%$ of nitrogen in given compound
$= \frac{28}{22400} \times \frac{V_{2}}{W} \times 100 = \frac{28}{22400} \times \frac{46.09}{0.35} \times 100$
$= 16.45 %$

JIPMER-2014 NEET-2011

Some Basic Concepts of Chemistry

228751 A mixture of ${CaCl}_{2}$ and $NaCl$ weighing $4.44 g$ is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to set $0.56 g$ of $CaO$ . The percentage of $NaCl$ in the mixture is (atomic mass of $Ca = 40$ )

1 75

2 31.5

3 40.2

4 25

Explanation:

Concerned reaction is.
${CaCl}_{2} + NaCl + {Na}_{2} {CO}_{3} \to {CaCO}_{3} + 2 NaCl + NaCl$
${CaCO}_{3} \overset{Δ}{\to} {CaO}_{(0.56 g)} + {CO}_{2}$
$100 {CaCO}_{3} = 111 g {CaCl}_{2}$
$1 g {CaCO}_{3} = 1.11 g {CaCl}_{2}$
$weight of {CaCl}_{2} = 1.11 g, weight of NaCl = 4.44 - 1.11 = 3.33 g$

2010

Some Basic Concepts of Chemistry

228752 An organic compound contains $60 % C; 4.48 %$ $H$ and $35.5 %$ O. Its empirical formula is

1

C_{9} H_{8} O_{4}

2

C_{5} H_{4} O_{2}

3

C_{5} H_{4} O_{4}

4

C_{9} H_{7} O_{6}

Explanation:

Element \% Atomic mass Moles
Carbon $60 12 \frac{60}{12} \Rightarrow 5$
Oxygen $35.5 16 \frac{35.5}{16} \Rightarrow 2.2$
Empirical formula $= C : H : O$
$= \frac{5}{2.2} : \frac{4.48}{2.2} : \frac{2.2}{2.2}$
$= 2.27 : 2.0 : 1.0$
$= 2.27 \times 4 : 2 \times 4 : 1 \times 4$
$= 9 : 8 : 4$
$= C_{9} H_{8} O_{4}$

TS-EAMCET (Engg.)

Some Basic Concepts of Chemistry

228753 If $0.765 g$ of an acid gives $0.535 g$ of ${CO}_{2}$ and $0.138 g$ of $H_{2} O$ then ratio percentage of $C$ and $H$ is

1

19 : 2

2

18 : 12

3

20 : 17

4

1 : 7

Explanation:

Given -
Mass of acid $= 0.765 gm$
Mass of ${CO}_{2} = 0.535 gm$
Mass of $H_{2} O = 0.138 gm$
$∴$ Acid $\times {CO}_{2} + H_{2} O$
$% C = \frac{12}{44} \times \frac{mass of {CO}_{2}}{mass of acid (compound)} \times 100$
$% C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 = 19.07 %$
$\begin{aligned} % H = \frac{2}{18} \times \frac{mass of H_{2} O}{mass of compound} \times 100 \\ % H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 = 2 % \end{aligned}$
So, the ratio of percentage of $C$ and $H$ is $19 : 2$ .

Shift-II

Some Basic Concepts of Chemistry

228748 $1 g$ of silver gets distributed between $10 {cm}^{3}$ of molten zinc and $100 {cm}^{3}$ of molten lead at $800^{\circ} C$ . The percentage of silver in the zinc layer is approximately

1 89

2 91

3 97

4 94

Explanation:

Let mass of $Ag$ in $Zn$ layer $= x$
$∴$ Mass of $Ag$ in lead layer $= 1 - x$
Concentration of Ag is zinc $= \frac{x}{10}$
$∴$ Concentration of $Ag$ in $Pb = \frac{1 - x}{100}$
$\frac{\frac{x}{10}}{\frac{1 - x}{100}} = 300$
$% v$ of $Ag$ in Zinc layer $= 97 %$

BCECE-2014

Some Basic Concepts of Chemistry

228749 In Dumas' method of estimation of nitrogen $0.35 g$ of an organic compound gave $55 mL$ . of nitrogen collected at $300 K$ temperature and $715 mm$ pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at $300 K = 15 mm$ )

1 15.44

2 16.45

3 17.45

4 14.45

Explanation:

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
Given,
Where, $p_{2} =$ pressure of $N_{2}$ at $STP = 760 mm$
$T_{2} =$ Temperature of $N_{2}$ at $STP = 273 K$
Volume of $N_{2}$ at STP (by gas equation)
$(\frac{P - P_{1}}{t + 273}) V_{1} \times \frac{273}{160} = V_{2}$
$P = 715 mm$ (pressure at which $N_{2}$ collected)
$P_{1} =$ aqueous tension of water $= 15 mm$
$T_{1} = t + 273 = 300 K$
$V_{1} = 55 ml$
$∴ V_{2} = \frac{(715 - 15) \times 55}{300} \times \frac{273}{760} = 46.098 mL$
$%$ of nitrogen in given compound
$= \frac{28}{22400} \times \frac{V_{2}}{W} \times 100 = \frac{28}{22400} \times \frac{46.09}{0.35} \times 100$
$= 16.45 %$

JIPMER-2014 NEET-2011

Some Basic Concepts of Chemistry

228751 A mixture of ${CaCl}_{2}$ and $NaCl$ weighing $4.44 g$ is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to set $0.56 g$ of $CaO$ . The percentage of $NaCl$ in the mixture is (atomic mass of $Ca = 40$ )

1 75

2 31.5

3 40.2

4 25

Explanation:

Concerned reaction is.
${CaCl}_{2} + NaCl + {Na}_{2} {CO}_{3} \to {CaCO}_{3} + 2 NaCl + NaCl$
${CaCO}_{3} \overset{Δ}{\to} {CaO}_{(0.56 g)} + {CO}_{2}$
$100 {CaCO}_{3} = 111 g {CaCl}_{2}$
$1 g {CaCO}_{3} = 1.11 g {CaCl}_{2}$
$weight of {CaCl}_{2} = 1.11 g, weight of NaCl = 4.44 - 1.11 = 3.33 g$

2010

Some Basic Concepts of Chemistry

228752 An organic compound contains $60 % C; 4.48 %$ $H$ and $35.5 %$ O. Its empirical formula is

1

C_{9} H_{8} O_{4}

2

C_{5} H_{4} O_{2}

3

C_{5} H_{4} O_{4}

4

C_{9} H_{7} O_{6}

Explanation:

Element \% Atomic mass Moles
Carbon $60 12 \frac{60}{12} \Rightarrow 5$
Oxygen $35.5 16 \frac{35.5}{16} \Rightarrow 2.2$
Empirical formula $= C : H : O$
$= \frac{5}{2.2} : \frac{4.48}{2.2} : \frac{2.2}{2.2}$
$= 2.27 : 2.0 : 1.0$
$= 2.27 \times 4 : 2 \times 4 : 1 \times 4$
$= 9 : 8 : 4$
$= C_{9} H_{8} O_{4}$

TS-EAMCET (Engg.)

Some Basic Concepts of Chemistry

228753 If $0.765 g$ of an acid gives $0.535 g$ of ${CO}_{2}$ and $0.138 g$ of $H_{2} O$ then ratio percentage of $C$ and $H$ is

1

19 : 2

2

18 : 12

3

20 : 17

4

1 : 7

Explanation:

Given -
Mass of acid $= 0.765 gm$
Mass of ${CO}_{2} = 0.535 gm$
Mass of $H_{2} O = 0.138 gm$
$∴$ Acid $\times {CO}_{2} + H_{2} O$
$% C = \frac{12}{44} \times \frac{mass of {CO}_{2}}{mass of acid (compound)} \times 100$
$% C = \frac{12}{44} \times \frac{0.535}{0.765} \times 100 = 19.07 %$
$\begin{aligned} % H = \frac{2}{18} \times \frac{mass of H_{2} O}{mass of compound} \times 100 \\ % H = \frac{2}{18} \times \frac{0.13}{0.765} \times 100 = 2 % \end{aligned}$
So, the ratio of percentage of $C$ and $H$ is $19 : 2$ .

Shift-II

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