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Solutions containing 23g HCOOH is / are

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46g of $\begin{array}{l} 70 % (\frac{w}{v}) H C O O H \\ (d_{s o l} = 1.40 g / m L) \end{array}$

50g of $\begin{array}{l} 10 M H C O O H \\ (d_{s o l} = 1 g / m L) \end{array}$

50g of $25 % (\frac{w}{w}) H C O O H$

46g of $5 M H C O O H (d_{s o l} = 1 g / m L)$

answer is A, B.

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Detailed Solution

$1) 46 g of 70 % \frac{w}{v} d = 1.40 g / m L$

$46 \times \frac{70}{140} = 23 g$

$2) M = \frac{w}{G M w} \times \frac{1000}{V} d_{s o l n} = g m / m L$

$10 = \frac{w}{46} \times \frac{1000}{50}$

$3) 25 g \to 100 g$

$\frac{25 \times 50}{100} \leftarrow 50 g$

$= 12.5 g$

$4) 5 = \frac{w}{46} \times \frac{1000}{46}$

$w = 10.58 g$

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