PartB 95 mL of a 070 M CsH1206 Express your answer using two

PartB 95 mL of a 0.70 M CsH1206 Express your answer using two significant figures. Submit Request Ans Part C 295 mL of a 180 MLiCI Express your answer using three significant figures. Submit Request An

Solution

Part B

first find number of moles of C₆H12O₆

volume of C6H12O6 used = 95 ml = 0.095 liter

molarity of  C6H12O6 = 0.7 molar

number of moles = molarity _X  volume in liter

number of moles C₆H₁₂O₆   = 0.7 X 0.095 = 0.0665 mol6

number of moles of C₆H₁₂O₆0.0665 mole

therefore 0.0665 mole C₆H₁₂O₆ = 0.0665 x 180.156 = 11.98 gram C₆H₁₂O₆

Mass of C₆H₁₂O₆ = 11.98 gram

Part c

295 ml = 0.295 liter

number of moles = molarity x volume in liter

number of mole of LiCl = 1.8 x 0.295 = 0.531

Number of mole of LiCl = 0.531 mole

molar mass LiCl = 42.394 gram/mole that mean one mole LiCl = 42.394 gram

0.531 mole LiCl = 0.531x42.394 = 22.51 grm LiCl

22.51 gram LiCl