Question

A 9.10-g sample of solid MnCl2·4H2O was heated such that the water turned to steam and was driven off. Assuming ideal behavior, what volume would that steam occupy at 1.00 atm and 100.0 °C?

Answer 1

Molar mass of MnCl₂.4H₂O is 521 g/mol

Molar mass of H₂O = 18g/mol

MnCl₂.4H₂O MnCl₂+4H₂O

From the above reaction,

1 mole of MnCl₂.4H₂O upon heating produces 4 moles of H₂O

                                           OR

521 g of MnCl₂.4H₂O upon heating produces 4x18 = 72 g of H₂O

9.10 g of MnCl₂.4H₂O upon heating produces M g of H₂O

M = ( 9.10x72) / 521

   = 1.258 g

Number of moles , n = mass/molar mass

                              = 1.258 g / 18(g/mol)

                              = 0.0698 mol

We know that PV = nRT

Where

T = Temperature = 100.0^oC = 100.0+273 = 373 K

P = pressure = 1.0 atm

n = No . of moles = 0.0698 mol

R = gas constant = 0.0821 L atm / mol - K

V = Volume of the steam = ?

V = nRT / P

    = 2.14 L

Therefore the volume of steam is 2.14L