In: Chemistry

# 26.33 g C6H12O6 in 1.26 L of solution 32.4 mg NaCl in 129.0 mL of solution...

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

## Solutions

##### Expert Solution

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

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

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