When 36.0 g of a solute having the empir...

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)`)

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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) ):

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - 0 . 34^(@) C . What is them molar ass of the substance ? ( K_(f) for water = 1. 86 K kg mol^(-1) )

18 g of a compound is dissolved in 10 kg of water so that the resulting solution freezes at - 8^(@) C . If K_(f) for water = 1 . 86 K/m . Calculate the molecular mass of the compound

35 g of compound is added in 1 kg of water so that the resulting solution freezes at - 10^(@) C . If K_(f) for water = 1. 86 K/m . Calculate the molecular mass of the compound

10g of non-volatile solute when dissolved in 100g of benzene raises its boiling point by 1^@ C. What is the molecular mass of the solute. ( k_b for benzene=2.53 K kg mol^-1 )

Calulate the molesules mass of a substance whose 7.0% by mass solution in water freezes at -0.93^(@)C . the cryosctopic constant of water is 1.86^(@)C kg mol^(-1) :

An aqueous solution boils at 101^(@)C . What is the freezing point of the same solution? (Gives : K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m )

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