18) D

Mr of water in ice = (1 x 2) + (16 x 1) = 18

Moles = Mass ÷ Mr

Moles = 6 ÷ 18 = ⅓ moles

Moles of steam would be the same = ⅓ moles

1 mole of steam at R.T.P. has a volume of 24 dm³ (24,000 cm³)

⅓ moles would have = ⅓ x 24,000 = 8,000 cm³

18) D

Mr of water in ice = (1 x 2) + (16 x 1) = 18

Moles = Mass ÷ Mr

Moles = 6 ÷ 18 = ⅓ moles

Moles of steam would be the same = ⅓ moles

1 mole of steam at R.T.P. has a volume of 24 dm³ (24,000 cm³)

⅓ moles would have = ⅓ x 24,000 = 8,000 cm³