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Standard XII

Chemistry

Types of Solution

Calculate the...

Question

Calculate the quantity of sodium carbonate (anhydrous ) required to prepare $250$ ml solution.

2.65

gram

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4.95

gram

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6.25

gram

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None of these

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Solution

The correct option is D $2.65$ gram
$m = \frac{n}{v} = \frac{m}{m \times V_{1} (l t)}$
$\Rightarrow x = \frac{m}{106 \times 25 \times 10^{- 3}}$
$\Rightarrow m = x \times 106 \times 25 \times 10^{- 3}$
$= 2.650$ gm

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Calculate the quantity of sodium carbonate (anhydrous ) required to prepare $250$ ml solution.

The correct option is D $2.65$ gram
$m = \frac{n}{v} = \frac{m}{m \times V_{1} (l t)}$
$\Rightarrow x = \frac{m}{106 \times 25 \times 10^{- 3}}$
$\Rightarrow m = x \times 106 \times 25 \times 10^{- 3}$
$= 2.650$ gm

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Calculate the quantity of sodium carbonate (anhydrous ) required to prepare 250ml solution.

The correct option is D 2.65 gramm=nv=mm×V1(lt)⇒x=m106×25×10−3⇒m=x×106×25×10−3=2.650gm

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EXAM PREPARATION

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FREE TEXTBOOK SOLUTIONS

STATE BOARDS

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Calculate the quantity of sodium carbonate (anhydrous ) required to prepare $250$ ml solution.

The correct option is D $2.65$ gram
$m = \frac{n}{v} = \frac{m}{m \times V_{1} (l t)}$
$\Rightarrow x = \frac{m}{106 \times 25 \times 10^{- 3}}$
$\Rightarrow m = x \times 106 \times 25 \times 10^{- 3}$
$= 2.650$ gm