Paired Exercise 14.23 How many milliliters of 0.879 M H3PO4 will contain the fol
ID: 1045061 • Letter: P
Question
Paired Exercise 14.23 How many milliliters of 0.879 M H3PO4 will contain the following? rrect. Did you convert moles to volume by multiplying the number of moles by the inverse of the molarity? (a) 0.29 mol H3PO mL H3PO4 the tolerance is +/-296 Incorrect. Did you first convert grams to moles and then multiply moles by the inverse of the molarity to determine the volume? (b) 25.9 9 H3PO mL H3PO4 the tolerance is +/-2% Incorrect. Did you use Avogadro's number to convert the number of molecules into moles? (c) 5.65 x 1022 molecules of H3PO L H3PO4 the tolerance is +/-2%Explanation / Answer
(a)
molarity = moles/volume
volume = volume in litres
moles of H3PO4 = 0.29 ( given in question )
molarity of H3PO4 = 0.879M ( given in question )
volume = 0.29/0.879
volume = 0.3299 Litres = 0.330 Litres
1 L = 1000 mL
0.330 L = 330 mL
volume = 330mL
( b )
moles = given mass / molecular mass
molecular mass of H3PO4 = 98 g/mol
given mass = 25.9g
mole = 25.9/98
mole = 0.264
again using the formula of molarity given above
volume = 0.264/0.879
volume =0.300 L = 300mL
(c)
no of molecules = moles * avagadros number
avagadros number = 6.023 * 1023
moles = no of molecules / avagadros number
moles = 5.65*1022 / 6.023*1023
moles = 0.094
molarity = moles / volume
volume = 0.094/0.879
volume = 0.107 L
volume 107mL
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