(A) Moles of first gas $n_1 = \frac{w_1}{m_1} = \frac{0.45}{60} = 0.0075 \ mol$. Moles of second gas $n_2 = \frac{w_2}{m_2} = \frac{0.22}{44} = 0.0050

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(A) Moles of first gas $n_1 = \frac{w_1}{m_1} = \frac{0.45}{60} = 0.0075 \ mol$. Moles of second gas $n_2 = \frac{w_2}{m_2} = \frac{0.22}{44} = 0.0050 \ mol$. Total moles $n_{total} = n_1 + n_2 = 0.0075 + 0.0050 = 0.0125 \ mol$. Partial pressure of the second gas $P_2 = \left( \frac{n_2}{n_{total}} \right) \times P_{total} = \left( \frac{0.0050}{0.0125} \right) \times 75 = 30 \ cm \ of \ Hg$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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If $0.45 \ g$ of a gas with molecular weight $60$ and $0.22 \ g$ of another gas with molecular weight $44$ are present,and the total pressure exerted by the mixture is $75 \ cm \ Hg$,then the partial pressure of the second gas is .......... $cm$ of $Hg$.

A
$30$
B
$20$
C
$10$
D
$40$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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If $0.45 \ g$ of a gas with molecular weight $60$ and $0.22 \ g$ of another gas with molecular weight $44$ are present,and the total pressure exerted by the mixture is $75 \ cm \ Hg$,then the partial pressure of the second gas is .......... $cm$ of $Hg$.

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$1 \ L$ of $N_2$ and $\frac{7}{8} \ L$ of $O_2$ gases are mixed at the same temperature and pressure. What is the relationship between their masses in the mixture?

An unknown chlorohydrocarbon has $3.55\%$ of chlorine. If each molecule of the hydrocarbon has one chlorine atom only,the number of chlorine atoms present in $1\,g$ of chlorohydrocarbon is (Atomic wt. of $Cl = 35.5\,u$; Avogadro constant $= 6.023 \times 10^{23}\,mol^{-1}$)

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$A$ cylinder contains a mixture of $5 \ g$ of $N_2$ and $6 \ g$ of $Ar$ gases. If the total pressure of the mixture of the gases in the cylinder is $30 \ bar$,then the partial pressure of $N_2$ gas is (in $bar$)

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