Chemistry, asked by Tirtharaj626, 1 year ago

What amount of nacl (in moles) is required to prepare 250 cm3 of a 0.200 mol dm3 solution?

Answers

Answered by KillerMS

26

Explanation:

V = 250 cm3

= 250 × 10^(-3) dm3

= 1/4 dm3

no. of mole = 0.2×1/4 mol

= 0.05 mol

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