In: Chemistry
26.33 g C6H12O6 in 1.26 L of solution
32.4 mg NaCl in 129.0 mL of solution
Calculate the Molarity in each, please explain
A)
Molar mass of C6H12O6 = 6*MM(C) + 12*MM(H) + 6*MM(O)
= 6*12.01 + 12*1.008 + 6*16.0
= 180.156 g/mol
mass of C6H12O6 = 26.33 g
we have below equation to be used:
number of mol of C6H12O6,
n = mass of C6H12O6/molar mass of C6H12O6
=(26.33 g)/(180.156 g/mol)
= 0.1462 mol
volume , V = 1.26 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 0.1462/1.26
= 0.116 M
Answer: 0.116 M
B)
Molar mass of NaCl = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol
mass of NaCl = 32.4 mg
= 0.0324 g [using conversion 1 g = 1000 mg]
we have below equation to be used:
number of mol of NaCl,
n = mass of NaCl/molar mass of NaCl
=(0.0324 g)/(58.44 g/mol)
= 5.544*10^-4 mol
volume , V = 129 mL
= 0.129 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 5.544*10^-4/0.129
= 4.30*10^-3 M
Answer: 4.30*10^-3 M