Mixture of two bivalent metals A and B having mass 2 ~g when dissolved in HCl at STP, 2.24 ~L H_2 is evolved. What is the mass of A present in mixture? (Atomic mass of A=15 u, B=30 u )

Correct answer: A. 1 ~g A+2 HCl arrow A Cl_2+H_2 Mole : (x)/(15) (x)/(15) B+2 HCl arrow BCl_2+H_2 Mole : (2-x)/(30) (2-x)/(30) Total moles of H_2=(x)/(15)+(2-x)/(30)=(2.24)/(22.4)=(1)/(10) x=1 ~g

AIIMS 201827 May

Exam: AIIMS
Year: 2018
Date: 27 May
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

AIIMS 2018 · 27 May · Chemistry · Some Basic Concepts of Chemistry

Mixture of two bivalent metals A and B having mass 2 ~g when...

Mixture of two bivalent metals

A

and

B

having mass

2 \mathrm{~g}

when dissolved in

\mathrm{HCl}

at STP,

2.24 \mathrm{~L}

\mathrm{H}_2

is evolved. What is the mass of

A

present in mixture? (Atomic mass of

A=15 \mathrm{u}, B=30 \mathrm{u}

)

Options

A $1 \mathrm{~g}$
B $1.5 \mathrm{~g}$
C $0.5 \mathrm{~g}$
D $0.75 \mathrm{~g}$

Correct answer

A. $1 \mathrm{~g}$

Step-by-step solution

\quad A+2 \mathrm{HCl} \rightarrow A \mathrm{Cl}_2+\mathrm{H}_2

Mole :

\frac{x}{15}

\frac{x}{15}

B+2 \mathrm{HCl} \rightarrow \mathrm{BCl}_2+\mathrm{H}_2

Mole :

\frac{2-x}{30}

\frac{2-x}{30}

Total moles of

\mathrm{H}_2=\frac{x}{15}+\frac{2-x}{30}=\frac{2.24}{22.4}=\frac{1}{10}

\Rightarrow \quad x=1 \mathrm{~g}

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