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CHEMISTRY(KCET)

285217 What amount of dioxygen (in gram) contains $1.8 \times$ $10^{22}$ molecules?

1 0.960

2 96.0

3 0.0960

4 9.60

Explanation:

(a) $6.022 \times 10^{23}$ molecules are present in 32 g of $O_{2}$
$∴ 1.8 \times 10^{22}$ molecules will be present in
$\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} g = 0.960 {g of O_{2}}^{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285218 0.30 g of an organic compound containing $C, H$ and O on combustion yields $0.44 g {CO}_{2}$ and $0.18 g H_{2} O$ . If one mol of compound weighs 60 , then molecular formula of the compound is

1

C_{3} H_{8} O

2

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

3

{CH}_{2} O

4

C_{4} H_{6} O

Explanation:

(b) Percentage of
$C = \frac{12}{44} \times \frac{0.44}{0.30} \times 100 = 40 %$
Percentage of $H = \frac{2}{18} \times \frac{0.18}{0.30} \times 100 = 6.6 %$
Percentage of $O = 100 - (40 + 6.6) = 53.4 %$
$Unknown environment 'tabular'$ & $Unknown environment 'tabular'$ \
\hline C & 40 & $\frac{40}{12} = 3.3$ & $\frac{3.3}{3.3} = 1$
\end{tabular}
Hence, empirical formula $= {CH}_{2} O$
$n = \frac{Molecular mass}{Empirical formula mass} = \frac{60}{30} = 2$
$\Rightarrow$ Molecular formula of the compound
$= {({CH}_{2} O)}_{2} = C_{2} H_{4} O_{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285219 $25 {cm}^{3}$ of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

1 0.064

2 0.045

3 0.015

4 0.032

Explanation:

(d) $(COOH)_{2} + 2 NaOH \to (COONa)_{2} + 2 H_{2} O$
No. of moles in 0.064 g of NaOH
$= \frac{0.064}{40} = 0.0016$
No. of moles of oxalic acid
$= \frac{0.0016}{2} = 8 \times 10^{- 4}$
Volume of solution (in L) $= \frac{25}{1000}$
Hence, molarity
$= \frac{No. of moles ofsolute}{Volume of solution (in L)}$
$= 8 \times 10^{- 4} \times \frac{1000}{25} = 0.032 M$

Karnataka CET 2014

CHEMISTRY(KCET)

285220 10 g of a mixture of BaO and CaO requires $100 {cm}^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately

(Given : molar mass of $BaO = 153$ )

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

(a) $BaO + 2 HCl \to {BaCl}_{2} + H_{2} O$
$CaO + 2 HCl \to {CaCl}_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
$= \frac{2.5 \times 100}{1000} = 0.25 moles$
Now, initial moles $= 4$ moles $(HCl)$
$+ 2 moles (BaO + CaO)$
Final moles of $HCl = 0.25$ moles
Final no. of moles of mixture of BaO and CaO
$= \frac{0.25 \times 2}{4} = 0.125 moles$
Let the mass of CaO be $x g$ and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0.125$
[Mol. mass of $CaO = 56, BaO = 153$ ]
$153 x + 56 (10 - x) = 0.125 \times 56 \times 153$
$153 x + 560 - 56 x = 0.125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511$
$\Rightarrow x = 5.268$
$∴ %$ of $CaO = \frac{5.268}{10} \times 100 = 52.68 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285217 What amount of dioxygen (in gram) contains $1.8 \times$ $10^{22}$ molecules?

1 0.960

2 96.0

3 0.0960

4 9.60

Explanation:

(a) $6.022 \times 10^{23}$ molecules are present in 32 g of $O_{2}$
$∴ 1.8 \times 10^{22}$ molecules will be present in
$\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} g = 0.960 {g of O_{2}}^{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285218 0.30 g of an organic compound containing $C, H$ and O on combustion yields $0.44 g {CO}_{2}$ and $0.18 g H_{2} O$ . If one mol of compound weighs 60 , then molecular formula of the compound is

1

C_{3} H_{8} O

2

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

3

{CH}_{2} O

4

C_{4} H_{6} O

Explanation:

(b) Percentage of
$C = \frac{12}{44} \times \frac{0.44}{0.30} \times 100 = 40 %$
Percentage of $H = \frac{2}{18} \times \frac{0.18}{0.30} \times 100 = 6.6 %$
Percentage of $O = 100 - (40 + 6.6) = 53.4 %$
$Unknown environment 'tabular'$ & $Unknown environment 'tabular'$ \
\hline C & 40 & $\frac{40}{12} = 3.3$ & $\frac{3.3}{3.3} = 1$
\end{tabular}
Hence, empirical formula $= {CH}_{2} O$
$n = \frac{Molecular mass}{Empirical formula mass} = \frac{60}{30} = 2$
$\Rightarrow$ Molecular formula of the compound
$= {({CH}_{2} O)}_{2} = C_{2} H_{4} O_{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285219 $25 {cm}^{3}$ of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

1 0.064

2 0.045

3 0.015

4 0.032

Explanation:

(d) $(COOH)_{2} + 2 NaOH \to (COONa)_{2} + 2 H_{2} O$
No. of moles in 0.064 g of NaOH
$= \frac{0.064}{40} = 0.0016$
No. of moles of oxalic acid
$= \frac{0.0016}{2} = 8 \times 10^{- 4}$
Volume of solution (in L) $= \frac{25}{1000}$
Hence, molarity
$= \frac{No. of moles ofsolute}{Volume of solution (in L)}$
$= 8 \times 10^{- 4} \times \frac{1000}{25} = 0.032 M$

Karnataka CET 2014

CHEMISTRY(KCET)

285220 10 g of a mixture of BaO and CaO requires $100 {cm}^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately

(Given : molar mass of $BaO = 153$ )

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

(a) $BaO + 2 HCl \to {BaCl}_{2} + H_{2} O$
$CaO + 2 HCl \to {CaCl}_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
$= \frac{2.5 \times 100}{1000} = 0.25 moles$
Now, initial moles $= 4$ moles $(HCl)$
$+ 2 moles (BaO + CaO)$
Final moles of $HCl = 0.25$ moles
Final no. of moles of mixture of BaO and CaO
$= \frac{0.25 \times 2}{4} = 0.125 moles$
Let the mass of CaO be $x g$ and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0.125$
[Mol. mass of $CaO = 56, BaO = 153$ ]
$153 x + 56 (10 - x) = 0.125 \times 56 \times 153$
$153 x + 560 - 56 x = 0.125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511$
$\Rightarrow x = 5.268$
$∴ %$ of $CaO = \frac{5.268}{10} \times 100 = 52.68 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285217 What amount of dioxygen (in gram) contains $1.8 \times$ $10^{22}$ molecules?

1 0.960

2 96.0

3 0.0960

4 9.60

Explanation:

(a) $6.022 \times 10^{23}$ molecules are present in 32 g of $O_{2}$
$∴ 1.8 \times 10^{22}$ molecules will be present in
$\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} g = 0.960 {g of O_{2}}^{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285218 0.30 g of an organic compound containing $C, H$ and O on combustion yields $0.44 g {CO}_{2}$ and $0.18 g H_{2} O$ . If one mol of compound weighs 60 , then molecular formula of the compound is

1

C_{3} H_{8} O

2

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

3

{CH}_{2} O

4

C_{4} H_{6} O

Explanation:

(b) Percentage of
$C = \frac{12}{44} \times \frac{0.44}{0.30} \times 100 = 40 %$
Percentage of $H = \frac{2}{18} \times \frac{0.18}{0.30} \times 100 = 6.6 %$
Percentage of $O = 100 - (40 + 6.6) = 53.4 %$
$Unknown environment 'tabular'$ & $Unknown environment 'tabular'$ \
\hline C & 40 & $\frac{40}{12} = 3.3$ & $\frac{3.3}{3.3} = 1$
\end{tabular}
Hence, empirical formula $= {CH}_{2} O$
$n = \frac{Molecular mass}{Empirical formula mass} = \frac{60}{30} = 2$
$\Rightarrow$ Molecular formula of the compound
$= {({CH}_{2} O)}_{2} = C_{2} H_{4} O_{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285219 $25 {cm}^{3}$ of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

1 0.064

2 0.045

3 0.015

4 0.032

Explanation:

(d) $(COOH)_{2} + 2 NaOH \to (COONa)_{2} + 2 H_{2} O$
No. of moles in 0.064 g of NaOH
$= \frac{0.064}{40} = 0.0016$
No. of moles of oxalic acid
$= \frac{0.0016}{2} = 8 \times 10^{- 4}$
Volume of solution (in L) $= \frac{25}{1000}$
Hence, molarity
$= \frac{No. of moles ofsolute}{Volume of solution (in L)}$
$= 8 \times 10^{- 4} \times \frac{1000}{25} = 0.032 M$

Karnataka CET 2014

CHEMISTRY(KCET)

285220 10 g of a mixture of BaO and CaO requires $100 {cm}^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately

(Given : molar mass of $BaO = 153$ )

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

(a) $BaO + 2 HCl \to {BaCl}_{2} + H_{2} O$
$CaO + 2 HCl \to {CaCl}_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
$= \frac{2.5 \times 100}{1000} = 0.25 moles$
Now, initial moles $= 4$ moles $(HCl)$
$+ 2 moles (BaO + CaO)$
Final moles of $HCl = 0.25$ moles
Final no. of moles of mixture of BaO and CaO
$= \frac{0.25 \times 2}{4} = 0.125 moles$
Let the mass of CaO be $x g$ and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0.125$
[Mol. mass of $CaO = 56, BaO = 153$ ]
$153 x + 56 (10 - x) = 0.125 \times 56 \times 153$
$153 x + 560 - 56 x = 0.125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511$
$\Rightarrow x = 5.268$
$∴ %$ of $CaO = \frac{5.268}{10} \times 100 = 52.68 %$

Karnataka CET 2014

CHEMISTRY(KCET)

285217 What amount of dioxygen (in gram) contains $1.8 \times$ $10^{22}$ molecules?

1 0.960

2 96.0

3 0.0960

4 9.60

Explanation:

(a) $6.022 \times 10^{23}$ molecules are present in 32 g of $O_{2}$
$∴ 1.8 \times 10^{22}$ molecules will be present in
$\frac{32}{6.022 \times 10^{23}} \times 1.8 \times 10^{22} g = 0.960 {g of O_{2}}^{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285218 0.30 g of an organic compound containing $C, H$ and O on combustion yields $0.44 g {CO}_{2}$ and $0.18 g H_{2} O$ . If one mol of compound weighs 60 , then molecular formula of the compound is

1

C_{3} H_{8} O

2

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

3

{CH}_{2} O

4

C_{4} H_{6} O

Explanation:

(b) Percentage of
$C = \frac{12}{44} \times \frac{0.44}{0.30} \times 100 = 40 %$
Percentage of $H = \frac{2}{18} \times \frac{0.18}{0.30} \times 100 = 6.6 %$
Percentage of $O = 100 - (40 + 6.6) = 53.4 %$
$Unknown environment 'tabular'$ & $Unknown environment 'tabular'$ \
\hline C & 40 & $\frac{40}{12} = 3.3$ & $\frac{3.3}{3.3} = 1$
\end{tabular}
Hence, empirical formula $= {CH}_{2} O$
$n = \frac{Molecular mass}{Empirical formula mass} = \frac{60}{30} = 2$
$\Rightarrow$ Molecular formula of the compound
$= {({CH}_{2} O)}_{2} = C_{2} H_{4} O_{2}$

Karnataka CET 2015

CHEMISTRY(KCET)

285219 $25 {cm}^{3}$ of oxalic acid completely neutralised 0.064 g of sodium hydroxide. Molarity of the oxalic acid solution is

1 0.064

2 0.045

3 0.015

4 0.032

Explanation:

(d) $(COOH)_{2} + 2 NaOH \to (COONa)_{2} + 2 H_{2} O$
No. of moles in 0.064 g of NaOH
$= \frac{0.064}{40} = 0.0016$
No. of moles of oxalic acid
$= \frac{0.0016}{2} = 8 \times 10^{- 4}$
Volume of solution (in L) $= \frac{25}{1000}$
Hence, molarity
$= \frac{No. of moles ofsolute}{Volume of solution (in L)}$
$= 8 \times 10^{- 4} \times \frac{1000}{25} = 0.032 M$

Karnataka CET 2014

CHEMISTRY(KCET)

285220 10 g of a mixture of BaO and CaO requires $100 {cm}^{3}$ of 2.5 M HCl to react completely. The percentage of calcium oxide in the mixture is approximately

(Given : molar mass of $BaO = 153$ )

1 52.6

2 55.1

3 44.9

4 47.4

Explanation:

(a) $BaO + 2 HCl \to {BaCl}_{2} + H_{2} O$
$CaO + 2 HCl \to {CaCl}_{2} + H_{2} O$
Reactions show that 2 moles of mix. of BaO and CaO requires 4 moles of HCl to react completely,
Hence, no. of moles of HCl
$= \frac{2.5 \times 100}{1000} = 0.25 moles$
Now, initial moles $= 4$ moles $(HCl)$
$+ 2 moles (BaO + CaO)$
Final moles of $HCl = 0.25$ moles
Final no. of moles of mixture of BaO and CaO
$= \frac{0.25 \times 2}{4} = 0.125 moles$
Let the mass of CaO be $x g$ and BaO is $(10 - x) g$
Then, $\frac{x}{56} + \frac{10 - x}{153} = 0.125$
[Mol. mass of $CaO = 56, BaO = 153$ ]
$153 x + 56 (10 - x) = 0.125 \times 56 \times 153$
$153 x + 560 - 56 x = 0.125 \times 56 \times 153$
$97 x = 1071 - 560$
$\Rightarrow 97 x = 511$
$\Rightarrow x = 5.268$
$∴ %$ of $CaO = \frac{5.268}{10} \times 100 = 52.68 %$

Karnataka CET 2014

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