Question

please answer both

Part B 27.03 g CH20. in 1.12 L of solution E AO M= 0.14 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Review your calculations; you may have made a rounding error or used Part 35.0 mg NaCl in 104.4 mL of solution 10 AED O ?

Answer 1

B)
Molar mass of C6H12O6,
MM = 6*MM(C) + 12*MM(H) + 6*MM(O)
= 6*12.01 + 12*1.008 + 6*16.0
= 180.156 g/mol


mass(C6H12O6)= 27.03 g

use:
number of mol of C6H12O6,
n = mass of C6H12O6/molar mass of C6H12O6
=(27.03 g)/(1.802*10^2 g/mol)
= 0.15 mol
volume , V = 1.12 L


use:
Molarity,
M = number of mol / volume in L
= 0.15/1.12
= 0.134 M
Answer: 0.134 M

C)
Molar mass of NaCl,
MM = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol


mass(NaCl)= 35.0 mg
= 0.035 g

use:
number of mol of NaCl,
n = mass of NaCl/molar mass of NaCl
=(3.5*10^-2 g)/(58.44 g/mol)
= 5.989*10^-4 mol
volume , V = 1.044*10^2 mL
= 0.1044 L


use:
Molarity,
M = number of mol / volume in L
= 5.989*10^-4/0.1044
= 5.737*10^-3 M
Answer: 5.74*10^-3 M