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