When 36.0 g of a solute having the empirical formula `CH_(2)O` is dissovled in 1.20 kg of water, the solution freezes at `-0.93^(@)C`. What is the moleculer formula of the solute ? (`K_(f) = 1.86^(@)C kg mol^(-1)`)
A
`C_(2)H_(4)O`
B
`C_(2)H_(2)O_(2)`
C
`C_(2)H_(4)O_(3)`
D
`C_(2)H_(4)O_(2)`
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The correct Answer is:
d
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When 36.0g of a non -volatile, non-electrolyte solution an empirical formula CH_(2)O is dissolved inn 1.20Kg of water. The solution freezes at -0.93^(@)C . What is the no. of oxygen atoms present per molecule of solute? K_(f) of H_(2)O=1.86K kg mol^(-1) , Freezing point of H_(2)O=273K
A solution containing 1.8 g of a compound (empirical formula CH_(2)O ) in 40 g of water is observed to freeze at -0.465^(@) C. The molecules formulea of the compound is ( K_(f) of water =1.86kg K mol^(-1) ):
(a) Define the following terms : (i) Mole fraction (ii) Ideal solution (b) 15.0 g of an unknown molecular material is dissolved in 450g of water . The resulting solution freezes at -0.34^(@)C . What is the molar mass of the material ? (K_(f) for water = 1.86 K kg mol^(-1))
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