29.2 % w / w H C l stock solution has a density of 1.25 g m L - 1 . The molecular weight of H C l is 36.5 g mo l - 1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M H C l is

Question

29.2 % w / w H C l stock solution has a density of 1.25 g m L - 1 . The molecular weight of H C l is 36.5 g mo l - 1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M H C l is,

MARKS App · Accepted Answer

Density of H C l = 1 . 25 g m L solution - 1 Weight of 100 mL solution = 125 g For a 29.2 % w / w H C l solution, weight = 125 100 × 29.2 = 36.5 gm So in 100 mL solution, moles of H C l = 36.5 36.5 = 1 moles So molarity of H C l = 10 M Using M 1 V 1 = M 2 V 2 Volume required = V 2 × 10 = 200 × 0.4 V 2 = 8 mL

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