Calculate the number of moles and the mass (in g) of the solute in each of the f

Question

Calculate the number of moles and the mass (in g) of the solute in each of the following solutions

(a) 2.40L of18.4MH2SO4, concentrated sulfuric acid

moles mol

mass g

(b) 153.0mL of4.3105MNaCN, a lethal concentration of sodium cyanide in blood serum

moles mol

mass g

(c) 5.22L of12.7MH2CO, the formaldehyde used to "fix" tissue samples

moles   mol

mass   g

(d) 330.mL of1.8106MFeSO4, a concentration of iron sulfate detectable by taste in drinking water

moles mol

mass g

I am having a hard time figuring this out and would appreciate the help.

Explanation / Answer

(a)

M = n/V

so

moles = volume x molarity

= 2.40L x 18.4 M

= 44.16 moles

1 mole of sulphuric acid = 98.07g

so

44.16 moles = 44.16 x 98.07

= 4330.77 g

so mass = 4330.77g

(b)

moles = volume x molarity

= 0.153 L x 4.3 105 M

= 0.6579 105 moles = 0.00006579 moles

1 mole of NaCN = 49 g

so

0.00006579 moles = 0.00006579 x 49

= 0.003223 g

so mass = 0.003223 g

(c)

moles = volume x molarity

= 5.22 L x 12.7 M

= 66.294 moles

1 mole of H2CO = 30.03 g

so

66.294 moles = 66.294 x 30.03

= 1990.80 g

so mass = 1990.80g

(d)

moles = volume x molarity

= 0.330 L x 1.8 106 M

= 0.594 106 moles = 0.00000594 moles

1 mole of FeSO4 = 151.90 g

so

0.00000594 moles = 0.00000594 x 151.90

= 0.0009 g

so mass = 0.0009 g

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