Question

Please answer the questions below to three significant figures.

a) What mass of KCl is required to make 44.0 mL of a  0.270 M KCl solution?
.......... g KCl
b) How many moles of potassium ions are present in the solution?
........ mol

Answer 1

al volume of Kel solution = 44.0mL ( 1000mL = IL) = 0.044 L concentration of kel solution - 0.270M moles of KCl a concentration (M) X volume (21 - 0.270M X 0.044L = 1:188 X10-2 mol molar may of ka = 74.55 g/mol ma 74.55 g/mol Ret = moles x molave mas = 1.188 x 10-2 mol X = 0.886 g Hence, 0.886 g of kel will be required

b) moles of KCl = 1.188 X10-3 moll Imole grues of ka kot lay) i mole of ktion K+(99) + (1-699) give 1.188 X 10 2 moles of ktions - 1188 x 10-2 mol of kel will Hence, 1:188 X 10-2 moles of Ktions present in the solution.
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