Question

A solution is prepared by dissolving 78 0 g of glucose (C6H1206 ) in 150 g of water. The final volume is 228 ml Calculate the Molarity (MM of glucose is 180 18 g/mol and MM of water is 18.02 g/mol) For units do not use fractional units like g/mol or g/mL QUESTION 23 A solution is prepared by dissolving 78.0 g of glucose (C6H1208) in 150 g of water. The final volume is 228 ml Calculate the Molality (MM of glucose is 180 18 g/mol and MM of water is 18 02 g/mol) For units do not use fractional units like g/mol or gimL QUESTION 24 A solution is prepared by dissolving 78.0 g of glucose (CH1206) in 150 g of water. The final volumes 228 ml Calculate the Mole Fraction (MM of glucose is 180.18 g/mol and MM of water is 18.02 g/mol) QUESTION 25 A solution is prepared by dissolving 78 0 g of glucose (CH1208) in 150 g of water. The final volume is 228 mt Cakulate the Mole percent (MM of glucose is 180. 18 g/mol and MM of water is 18.02 g/mol) Do not include percent sign. Only write the number

Answer 1

P. A solution is prepared by dissolving 78.0g of glucose (C6H1206) in 150g of waters. The final rolume is 228mh. Molarity of the solution=? * The amount of glucose added a mass of glucose dissolved mofaromass of glucose 78.09 - Mefaritya amount inmole 180-18gmots Hy volume inlitne - (78.0 mole. 3 1180.18) mole. G Tenia) ms 0.2282 = 1.89869 (M) ~ 1.899(M) 9. 23 Total number of moles of solate contained by 1 kg (looog) of Sevent, is called the mola concentrations aro Molality of the solution. ... Tofal members of moles of glucare present In 150g of water a 780g . .. 180. 18 g/mol "Teorie) mole n. 150g watero ContainG > 78 moles of glucose, 278_ 1000) mole gluese 18018 | 1507 52.886 moal e ? 2.886 cm) :: Mofality of the solution - 2-8866m). e looogº..."

119.24 Here total moles of glucose present in the medium (Pectivos) 78009 fte male 112007 180118gmatt 180-18).*** tofal ons les of Hyo present (MH50) > 150g 18:02 glonal (150 mole. ose ". Mole fraction of glucose &CH2O6 MCGH12O6 MCGHEO6th H2O 2/78 i 180.18) - - 0.0494 17818) + ( 18.02) 150 l sactio E role fraction of H20 (*tho) - MH2O > (1962) nazo + MCGH1206 = 0.950 2.25 tatand molar of the selection (n) Nchipet Whyo -. 79 8.757 mde (180-18) + (150) 20.4329+8.32447 móle 18.02) 910.2ngid

in i, mole pencent of glucose.: MC6H12O6 _ x 100 > () 8.757 20.4329 x100 24094 .8757 -x100

1 And, the mole percent of waters a milho x100 / 150 l 18.02 x 100 80757 28.3241 x 100 =95.05 8.757