yahyatahan yahyatahan

24-02-2022
Chemistry

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What amount of nacl (in moles) is required to prepare 250 cm3 of a 0.200 mol dm-3 solution?

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rubenchhatkuli rubenchhatkuli

24-02-2022

Answer:

V = 250 cm3

= 250 × 10^(-3) dm3

= 1/4 dm3

no. of mole = 0.2×1/4 mol

= 0.05 molExplanation:

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