29.2 \%(w / w) HCl stock solution has a density of 1.25 gmL^ -1 . The molecular weight of HCl is 36.5 ~g ~mol^ -1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is :

Correct answer: 0 Molarity of stock solution of HCl =(29.2 × 1000 × 1.25)/(100 × 36.5) Let the volume of stock solution required =V ~mL Thus, V × (29.2 × 1000 × 1.25)/(100 × 36.5)=200 × 0.4 V=8 ~mL

JEE Advanced 2012Paper 1

Exam: JEE Advanced
Year: 2012
Paper: Paper 1
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Advanced 2012 · Paper 1 · Chemistry · Some Basic Concepts of Chemistry

29.2 \%(w / w) HCl stock solution has a density of 1.25 gmL^ -1 ....

29.2 \%(\mathrm{w} / \mathrm{w}) \mathrm{HCl}

stock solution has a density of

1.25 \mathrm{gmL}^{-1}

. The molecular weight of

\mathrm{HCl}

36.5 \mathrm{~g} \mathrm{~mol}^{-1}

. The volume

(\mathrm{mL})

of stock solution required to prepare a 200

\mathrm{mL}

solution of

0.4 \mathrm{M} \mathrm{HCl}

is :

Correct answer

Step-by-step solution

Molarity of stock solution of

\mathrm{HCl}

=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}

Let the volume of stock solution required

=V \mathrm{~mL}

Thus,

V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4

\Rightarrow V=8 \mathrm{~mL}

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