Question

hass of KHCgHO4 g/mole 2231 22.4345 2323 AN 21.7362 26,144 a Sample # 1 3 Masses of KHC,H,O, Samples Mass of bottle +sample 2 2, 2380 22.2B2 21.7367 236 20.7 20.3760 0LES 0.501 0.111 0366 Mass of bottle + remaining material Mass of sample SS.49 Buret Readings 7-50 8S 17.90 135 Final reading mL O,00 0.00 05o49 mL Initial reading 50.64l 250ml 190m 6.845)m Volume NaOH added Calculation of Molar Concentration 0.SI 0.999 0346 O.ds Moles KHCgH2O4 mol Moles NaOH 50.64 256ml 1a6 645 MBT03210.27 O.0435 0227 Volume NAOH added Molarity NaOH 0.2127 a213 M erage molarity NaOH Y03.50 ppt cision as RAD

can you help me fibd moles of NaOH

Answer 1

we know molarity = moles / liter = moles / 1000 mL

we have volume in mL & molarity in M

so ==>[ moles / 1000 mL] x mL = moles ( mL gets cancelled)

trial-1:

moles NaOH = NaOH molarity x volume of NaOH = 0.0435 M x 50.69 mL = 0.0022 moles

trial-2:

moles NaOH = NaOH molarity x volume of NaOH = 0.3277 M x 7.50 mL = 0.0025 moles

trial-3:

moles NaOH = NaOH molarity x volume of NaOH = 0.2716 M x 17.90 mL = 0.0049 moles

trial-4:

moles NaOH = NaOH molarity x volume of NaOH = 0.2627 M x 6.8451 mL = 0.0018 moles

NOTE: Please change volume NaOH added values to Liters because you entered in mL, i mean 50.64 should be 0.05064 L

1 L = 1000 mL

Hope this helped you!

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