The correct option is C 8 mL Molarity of stock solution (soln.) of HCl =(% mass)(Density of solution) (10)(Molar mass) (29.2)(1.25)(10)36.5=10M ...

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Gay Lussac's Law

29.2

Question

29.2% (w/W) HCl stock solution has a density of 1.25 $g m L^{- 1}$ . The molecular weight of HCl is 36.5 $m o l^{- 1}$ . The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is:

5 m L

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6 m L

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8 m L

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15 m L

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Solution

The correct option is C $8 m L$
Molarity of stock solution (soln.) of HCl
$= \frac{(% m a s s) (D e n s i t y o f s o l u t i o n) (10)}{(M o l a r m a s s)}$

$\frac{(29.2) (1.25) (10)}{36.5} = 10 M$

According to dilution equation $M_{i} V_{i} = M_{f} V_{f}$
$V_{i} = \frac{M_{f} V_{f}}{M_{i}} = \frac{(0.4 M) (200 m L)}{(10 M)} = 8 m L$ .

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Similar questions

29.2 % (w / w) H C l

1.25 g m L^{- 1}

H C l

36.5 g m o l^{- 1} .

(m L)

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1.5 g m L^{- 1} .

36.5 g m o l^{- 1} .

1.5 g m L^{- 1} .

36.5 g m o l^{- 1} .

1.25 g m L^{- 1} .

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What volume of 29.2% w/w HCl solution of density 1.25 g/ml is required to prepare 200 ml of 0.4 M HCl solution.

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