Question

How many chloride ions are present in 250.0 mL of 0.173 M CaCl2 solution?

How many mL of 1.55 M NaOH solution do you need to prepare 125.0 mL of 0.0500 M NaOH solution?

Answer 1

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Moles of D.173 MX 0.250 L o.ou3 mol. - Moles of ce = call ax moles of 2x 01043 nol 0.086 mol No. of cé longa - 2 6.02 x1025 x 0.086 5.2x1022 longe - using 25 MiVIE M2V2 1.55 4x Via 125. Omel x 0 osaomh 2) Vi= 4.03 ml i Vinumele 4ml of solution is required

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